If you are coming to this table of contents now for the first time, it may seem as though a black belt in Geometry–Do is a mountain too high to climb. But I tell you, the next two years are going to pass anyway, so why not go for the gold? It is an accomplishment you can boast about for the rest of your life. What else can you do as a teenager that you will be proud of as an old man?
The most fundamental question of philosophy is what it means to be human. We used to think that tool making was the defining characteristic until Jane Goodall observed chimpanzees dipping for insects with purposemade twigs. Stone is harder to work, but that does not make a man. I argue that the defining characteristic is abstract reasoning using symbols to represent things. If I hide a child’s toy and then show her a drawing of the floor plan of our house and say, “This is our house, here is your room, here is the kitchen and X marks the spot where your toy is hidden,” she will look at the drawing and then go get her toy. This would not work with our dog. I could draw maps on the floor every day and he would die of old age without ever showing a glimmer of understanding ^{1}. It is geometry, not engineering, that distinguishes us from the animals.
Immanuel Kant spoke of Euclidean geometry and is widely derided today by people who feel that the work of Lobachevski and Bolyai consign Critique of Pure Reason to the rubbish bin. But at the time of publication, 1781, there was only one geometry; the work of Lobachevski and Bolyai came 50 years later and Riemann 25 years after that. What matters is not that Kant used the term Euclidean as though it were synonymous with geometry but that he cited only theorems before Book I, Proposition 28 of The Elements, the first theorem to use Euclid’s fifth postulate.^{2}
What Kant really meant is that you do not have to explain to children that a segment being straight implies that it is the shortest path between two points, or that the points on the shortest path are between the two points. Basic concepts like what points and lines are and what it means for a line to be straight or for a point to be between two other points do not require explanation; the geometry teacher is just assigning names to concepts that the child already understands. This is the same point that the Epicureans made when they scoffed at the triangle inequality theorem for being evident even to an ass (donkey), who knows what the shortest path to a bale of hay is^{3}.
Some geometers define a segment as two points and all those between them. But the shortest path and the points between are the same thing. A point would not be thought of as between two others if it were not on the shortest path from one to the other. Defining a segment as the shortest path between two points is more useful because there are lots of problems in geometry about minimizing lengths; e.g. problems 3.2 to 3.4 and the problems of Fermat and Fagnano. Our axioms are chosen not just to be intuitive to a child, but also to be productive for the adults.
This textbook is intended for both general and honor students in both America and India. It is in American English, but without the condescending tone prevalent in American textbooks. Military schools must require a twoyear program to have time for machine gun emplacement.
White and yellow belt serve three purposes: It provides a theoretical foundation for the later chapters; it prepares students for a class in nonEuclidean geometry; and it teaches construction workers what they need to know to excel in carpentry, masonry, concrete, and asphalt work. The first two are achieved by teaching only absolute geometry; that is, the theorems common to both Euclidean and nonEuclidean geometry. Orange belt introduces the parallel postulate. This may seem a bit abstract for construction workers, but the difference between intrinsic and metric geometry is illustrated by solving the same problem with a pair of 100’ chalked strings or a 25’ tape measure, so it is often the construction workers who are first to accept intrinsic geometry.
Everybody – American, Indian, general and honor students – should take a full semester going through white and yellow belt, proving every theorem and solving every problem. For the Americans, this is 10th grade, and for the Indians, 8th grade; the stronger students will have seen many, but not all, of these theorems before, though probably not treated so formally, and the weaker students will appreciate that the accompanying website gives stepbystep explanations.
High School Shop Class: Everything you need to know is white and yellow belt. Stretch it out to a full year by studying geometry three days a week and memorizing the building codes or doing field work at a construction site two days a week. Award an orange belt on completion.
American general students: Spend 2nd semester memorizing all those theorems labeled R (Russian 7th, 8th and 9th grade) in the index. Spend the last few weeks on How to be a Tetrahedonist and Squares and Rectangles and Rhombi! Oh My! and How to Take Standardized Exams that Define Geometry in Terms of Motion. Award a green stripe on their orange belts.
American honor students and all Indians: Spend 2nd semester on orange belt; award a green belt on passing the greenbelt entrance exam. Yellow and orange are shades of white, but green is intermediate! If students pass the class, but not the exam, a green stripe on their orange belts.
Secondyear geometry: Review orange belt and then have those with a green stripe on their orange belts take the green belt exam on add/drop day. If they fail a second time, then goodbye. Award a red belt to those who complete green belt. The exit exam requires either going to sea and plotting a course with a sextant or laying some machine guns to defend one’s town.
I received 72 credits in the math department as an undergraduate, which was every class that the university offered. But even after all those classes, I can say without exaggeration that high school geometry was the most enjoyable math class I ever took. Actually, inspiring is probably a better word because it was geometry that made me decide to become a mathematician.
Why did I find geometry so inspiring? On reflection, I believe that this is because I am basically an inventor at heart. Indeed, I have invented a number of things since graduating, including an economic theory, a cryptosystem, a system to play casino blackjack and a weapon to kill SAM crews on a skyscraper without exposing aircraft to missile fire and without risk to civilians in the streets below. Other math classes taught me a lot of formulas and equations, but only geometry gave me the sense of inventing a selfcontained science from the ground up, which is really what it is like – I can tell you from experience – to invent some new weapon or device.
Sadly, since becoming a grown up, no conversation I have had with a high school student or recent graduate has indicated that they felt this sense of invention when studying geometry. Without exception, they felt that geometry was a big waste of time, a class stuck incongruously between Algebra I and Algebra II that – in actual practice – amounted to nothing more than a review of Algebra I, a subject that they had already demonstrated their mastery of by acing. It is my hope that this textbook will help kids discover the sense of invention that so thrilled me!
A word to the wise, Grasshopper: In the military they say, “train like you fight.” The website that accompanies white and yellow belt has illustrations but, after that, you must draw your own. And you are advised to draw your own in white and yellow belt before consulting the website. Students fail because they get too relaxed when studying. Their only movement is their eyeballs traversing back and forth. They say “I understand” after reading each proof, but their mind is wandering. Then they fail the exam and say, “I’m bad at exams.” No, they are bad at studying.
You would not last long in the ring if your only experience with boxing was sitting on your butt watching Monday Night Fights. And, when I write a geometry problem on the board, set a blank sheet of paper on your desk and say “Solve it!” it had better not be the first time you have used your compass and straight edge. You wouldn’t go into combat with a rifle you have never fired at the range, would you? These are not going to be those pansy exams where they solve the entire problem but leave one theorem citation blank and asks you to fill it in. We just state the problem and hand you a blank sheet of paper. It is up to you to write something intelligent on it.
Good luck! Less than one in ten pass the greenbelt entrance exam; try to be that elite student!
\(\overline{EF}\) is given to be a third the length of \(\overline{EG}\). Which of these answers is true?
1.\(\overline{EI}=\frac{1}{2}\overline{IG}\), 2. \(\overline{EI}=\frac{2}{3}\overline{IG}\) , 3. \(\overline{EI}=\frac{3}{4}\overline{IG}\), 4.\(\overline{EI}=\overline{IG}\)
They are the same length! It is an optical illusion. The Note to Philosophers claims that the terms shortest path, straight, inside and between are intuitive. But I never said that it is a good idea to use intuition to compare lengths, angles or areas. Good luck on your journey, Grasshopper!
How to Be a Tetrahedonist
The Varsity Tutors Advanced Geometry Exam has nothing to do with geometry and everything to do with memorizing obscure algebra formulas that you will never use^{4}. Varsity Tutors are called tetrahedonists because most of their formulas are about tetrahedrons^{5}. A useless shape!
is a sphere’s radius; also, is a cylinder’s or a cone’s base radius. is a tetrahedron’s edge.

sphere 
cylinder 
cone 
Tetrahedron 
height 

\(h\) 
\(h\) 
\(\frac{\sqrt{6}}{3}x\) 
slant length 


\(s=r^{2}+h^{2}\) 

base area 

\(B=πr2\) 
\(B=πr2\) 
\(B=\frac{\sqrt{3}}{4}x^2\) 
lateral area 

\(L=2πrh\) 
\(L=πrs\) 

total area 
\(A=4πr^{2}\) 
\(A=2B+L\) 
\(A=B+L\) 
\(A=\sqrt{3}x^{2}\) 
volume 
\(V=\frac{4}{3}πr^{3}\) 
\(V=Bh\) 
\(V=\frac{Bh}{3}\) 
\(V=\frac{{\sqrt{2}}}{12}x^{3}\) 
For kites and rhombi (a rhombus is a kite with equal sides),\(A=\frac{pq}{2}\) with p and q the diagonals. The sides of a rhombus are \(x=\frac{\sqrt{p2+q2}}{2}\) . Also, you must know that a trapezoid’s area is \(A=hw\) with \(w\) the semisum of the base and the top. The rest of the exam is basic algebra: To reflect a polynomial over \(x=0\) , negate the odd terms; to reflect it over \(y=0\) , negate all the terms. Know how to complete the square to get \(r2=(xx0)^{2}+(yy0)^{2}\) and the meaning of \(x0,y0,r\). Given two points, be able to find the line through them, the distance between them, and their midpoint. Also, find a line through a point given its slope or given a perpendicular line. Know the definitions of sine, cosine and tangent, but you will not be asked about trigonometric identities.
Isn’t that amazing? Without knowing a single geometry theorem, you can now call yourself a Varsity Tutor master of “advanced geometry.” But, when the glow of this heroic achievement wears off, you might want to read Geometry–Do. You will learn how to make a house square, navigate a ship at sea, lay .50 cal. machine guns, survey farmland and all kinds of cool stuff!
Why Do Americans Stink at Math?
At the end of the book in the humor section I photocopy a newspaper article that states that our local community college administers an entrance exam for incoming students and finds that 90% require remedial algebra and 80% require remedial prealgebra before they can be allowed to take any actual collegelevel classes. Morons! How could our schools have failed so miserably?
• America is unique in spending too much time in high school and too little in middle school.

Elemnetary 
Middle 
High school 
America Grade 
1st to 6th grade 
7th and 8th grade 
9th to 12th 
India Grade 
1st to 5th grade 
6th to 10th grade 
11th and 12th 
Russia Grade 
1st to 4th grade 
5th to 9th grade 
10th and 11th 
Why does America have a twoyear middle school and a fouryear high school? Sports. Kids can only master a sport if they spend four years under the same coach. In Russia and India, the highest honor is not varsity sports but being among the elite few who graduate.
• America is unique in having such large high schools. Mexican students tell me that they are about two years ahead of American students in math. But they also complain of being intimidated by the campuslike atmosphere of American high schools^{6}. Our high schools have thousands of students while theirs have hundreds. Why so big? A high school of two thousand students is a football powerhouse while one of two hundred students must scrape the bottom of the barrel to get thirty boys for their varsity football team.
• Common Core institutionalizes mediocrity. Let’s all move to Lake Wobegon where no child is left behind and every student succeeds! Unfortunately, India has four times our population and a meritocracy. While every Indian child can complete high school, they expect twothirds attrition. The result is that every Indian with a high school diploma is smarter than even our top students, and there are more of them than all of our students.
• A false dichotomy of either bureaucrats or housewives dictating curriculum. Neither are qualified for this task. The professors in a community college or a university should write the entrance exam for their own school. The nearby high schools should have three tracks to train students to take the exam needed to enter a trade school, a state university or an elite private college. Accept that there is attrition that will lead to flipping hamburgers.
What Should Replace Common Core?
Common Core proponents sneer^{7} at their opponents for having no goal or some goof ball goal.
• A few Common Core opponents are religious zealots that are still trying to deny evolution.
• Most of them are housewives that think they know it all when it comes to elementary education, but they become mysteriously silent when highschool education is discussed.
In sharp contrast, the proponents of Common Core know exactly what “defeat” would mean: a new paymaster – some billionaire other than Bill Gates trying to bribe his way into a monopoly on educational software. But nobody is as rich as Bill Gates, so they know this is not happening.
One cannot help but notice that the group whose opinions are never sought are the university professors that will be receiving these socalled “college ready” students. Indeed, content experts, as they are called, have about as much say regarding the curriculum as the high school janitor does. This point is often concealed by Common Core proponents’ boastful talk of their PhDs. But their PhDs are not in the subjects that they are teaching; they are PhDs in education^{8}.
Who controlled the curriculum in the 19th century? George Wentworth (1899, p. 180) writes:
Proposition XXIII The square on the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side of the triangle.
Proposition XXIV In any triangle the product of two sides is equal to the product of the diameter of the circumscribed circle by the altitude upon the third side.
This theorem may be omitted without destroying the sequence. Props. XXIII and XXIV are occasionally demanded in college entrance examinations, but they are not necessary for proving subsequent propositions or for any of the exercises. Teachers may therefore use their judgement as to including them.
Geometry–Do proves both theorems, but the point is that 19th century professors controlled the highschool curriculum by writing entrance exams. Sadly, such entrance exams no longer exist.
Euclid’s Postulates Plus One More
Segment  Two points fully define a segment. 
Triangle  Three points fully define a triangle. 
Line  A segment fully defines a line. 
Circle  The center and the radius fully define a circle. 
Right Angle  All right angles are equal to each other. 
Parallel  A line and a point not on it fully define the parallel through that point. 
The postulates are in terms of fully defined, which means that a figure with the given characteristics is unique, if it exists. SSS fully defines a triangle, but it may not exist if one side is too long; ASS does not fully define a triangle. Under defined means figures with the given characteristics are legion; more information is needed. John Playfair stated the parallel postulate roughly as I and David Hilbert do, which can be proven to be equivalent to Euclid’s Fifth Postulate.
If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
While Hilbert and I both found Euclid’s postulate to be convoluted and chose Playfair’s version, and we both reject real numbers as unsupported by our postulates, we otherwise took separate paths. Geometry–Do is like Hilbert’s geometry, but it is unique and has its own postulates.
Euclid also had five “common notions,” which vaguely describe what modern mathematicians call equivalence relations, total orderings and additive groups.
Equivalence Relations and Total Orderings
A relation is an operator, R , that returns either a “true” or a “false” when applied to an ordered pair of elements from a given set. For instance, if the set is integers and the relation is equality, then \(5=5\) is true, but \(5=4\) is false. Relations must be applied to objects from the same set. For instance, \(\overline{EF} = \angle G \) is neither true nor false; it is incoherent. There are four ways that relations may be characterized. It is never true that a relation has all four, but some have three.
Transitive 
a R b and b R c implies a R c 
Reflexive 
a R a 
Symmetric 
a R b implies b R a 
AntiSymmetric 
a R b and b R a implies a= b 
A relation that is transitive, reflexive and symmetric is called an equivalence relation. The equivalence relations considered in geometry are equality ,=, which applies to segments, angles or areas; congruence ,\(\cong\), which applies to triangles; similarity,\(\sim\) , which applies to triangles; and parallelism, \(\), which applies to lines. \(\overleftrightarrow{EF}\overleftrightarrow{GH}\) means that \(\overleftrightarrow{EF}\) and \(\overleftrightarrow{GH}\) do not intersect.
Since segments are known only by their length, \(\overline{EF}== \overline{GH}\) means that \(\overline{EF}\) and \(\overline{GH}\) are the same length. It does not mean that they are the same segment; they may be in different locations. Since length is the same regardless of direction, it is always true that \(\overline{EF}= \overline{FE}\). But triangles are known, not by just one magnitude, but by six. The vertices are ordered to show which ones are equal. \(\overline{EFG}\cong\overline{HIJ}\) means that \(\overline{EF}=\overline{HI}\), \(\overline{FG}=\overline{IJ}\) ,\(\overline{GE}=\overline{JH}\) ,\(\angle E\cong\angle H\) ,\(\angle F\cong\angle I\) and \(\angle G\cong\angle J\). Beware! Writing the vertices of a triangle out of order is one of the most common mistakes made by beginning geometers, and it is always fatal to a proof.
A quadrilateral is a union of two triangles; congruence or similarity holds if and only if both pairs of triangles are congruent or similar. If \(\overline{EFG}\cong \overline{IJK}\) and \(\overline{EHG}\cong \overline{ILK}\), then, \(\overline{EFHG}\cong \overline{IJLK}\). Analogously, if \(\overline{EFG}\sim \overline{IJK}\) and \(\overline{EHG}\sim \overline{IKL}\), then, \(\overline{EFGH}\sim \overline{IJKL}\). Similarity is defined as two triangles with all corresponding angles equal, so \(\overline{EFG}\sim \overline{IJK}\) and \(\overline{EHG}\sim \overline{ILK}\) means that six pairs of corresponding angles are equal. This is more than just saying that the four corresponding interior angles of \(\overline{EFGH}\) and \(\overline{IJKL}\) are equal; thus, it is not true that proving these four equal is sufficient to prove \(\overline{EFGH}\sim \overline{IJKL}\). A counterexample is a square and a rectangle; they both have all right angles, but they are not similar. This is one reason why we do not define quadrilaterals as foursided figures. This is a vacuous definition that has led many beginners to err by claiming that a square and a rectangle are similar. Also, our definition makes quadrilaterals a continuation of triangles; American schools have these as semester programs that can be taken in either order.
A relation that is transitive, reflexive and symmetric is an equivalence relation and there are four in geometry: equality, congruence, similarity and parallelism. Relations that are antisymmetric can only be defined if we have already defined equality, because equality is referenced in its definition. A relation that is transitive, reflexive and antisymmetric is called an ordering. Geometers only consider one: less than or equal to,\(\leq\) . An ordering is total if \( a \leq b \) or \( b \leq a \), always. A set with both an equivalence relation,= , and a total ordering, \(\leq\), is called a magnitude. There exist orderings that are not total, such as subset, but these are not used in geometry. Less than, \(<\) , means \(\leq\) but not =. It cannot be defined until both \(\leq\) and \(=\) have been defined.
Note that our definition of magnitude does not imply that real numbers can be associated with lengths, angles or areas; only that the relations \(=\) and \(\leq\) exist and have the required properties. It does imply that magnitudes are unique, which is what the replication axiom below is stating.
Equal magnitudes are an equivalence relation and can be reproduced wherever needed; that is, compasses do not collapse when lifted from the paper but are like holding a rope at a length. Compasses that collapse would be like surveyors who can walk a rope around an arc but, the moment the center guy takes a step, their rope turns to smoke. Because errors accumulate, it is not possible to put hash marks every foot – a quarterinch error in every mark is an error of several feet per hundred yards – plus shrinkage or expansion as temperature and humidity change. This is why we use a straight edge, not a ruler; but the idea that a compass cannot be lifted off the paper to mark a length elsewhere makes geometry a parlor game, not a science.
An equivalence class is a set of objects that are equal, congruent, similar or parallel to each other. Equivalence classes can be defined in reference to an existing equivalence class. For instance, if an equivalence class is defined as all the angles equal to a given angle, then all the angles complementary to any member of that class are equal to each other; that is, they form their own equivalence class. All the angles supplementary to any member of that class are also equal to each other. If an equivalence class is defined as all the lines parallel to a given line, then all the lines perpendicular to any member of that class are parallel to each other. All the circles with radii equal to any member of an equivalence class of equal segments are an equivalence class.
Equivalence also refers to statements that can be proven if the other one is assumed, and in either order. For instance, Euclid’s fifth postulate and Playfair’s postulate are equivalent because, assuming either to be true, it is possible to prove that the other is true. The equivalence of theorems can be expressed by separating them with the phrase “if and only if,” which can be abbreviated “iff.” Proof in the other direction is called the converse; that is, if \(X\) implies \(Y\), then the converse is that \(Y\) implies \(X\). If \(X\) and \(Y\) are equivalent, then both implications are true.
Proof by contradiction when there is only one alternative that must be proven impossible is called a dichotomy. A trichotomy (e.g. ASA congruence) has three alternatives. A magnitude can either be less than, equal to or greater than another, and only one of these three is desired; thus, by proving the other two to be impossible, we know that it is the one that makes the theorem true.
We define an additive group as a set and an operation (addition) that has these properties:
Associative property 
\((a+b)+c=a+(b+c)\) 
Commutative property 
\(a+b=b+a\) 
Existence and uniqueness of an identity 
\(a+0=a=0+a\) 
Existence of inverses (identity is its own) 
\(a+(a)= 0 =(a)+a\) 
There exist magnitudes that are not additive groups, such as economic value. Given a choice between \(a\) or \(b\), it is always possible for a person to choose one above the other. But, because \(a\) may substitute for or be a complement to \(b\), they are not independent the way geometric magnitudes are. There are also additive groups that cannot be ordered, such as matrices. Matrices of the same dimension are an additive group, but we cannot say \(a \leq b\) for any two.
On the first day of class I ask the students to look back to a time eight or ten years prior, when they were little kids and knew only how to add and subtract; multiplication and division was still scary for them. I assure them that geometry will be like going back to 1st grade. Sticking segments together end to end or angles together side by side is no more difficult than 1st grade problems about adding chocolates to or subtracting chocolates from a bowl of candies. How easy is that?
Replication Axiom
Given \(\overline{EF}\) and \(\overrightarrow{E'G'}\), there exists a unique point F' on \(\overrightarrow{E'G'}\) such that \(\overline{EF}\)=\(\overline{E'F'}\).
Given \(\angle EFG\) and \(\overrightarrow{F'E'}\), on a side of \(\overrightarrow{F'E'}\) there exists a unique ray \(\overrightarrow{F'G'}\) such that \(\angle EFG\)=\(\angle E'F'G'\).
Interior Segment Axiom
If \(G\) is between \(E\) and \(F\) then \(\overline{EG}<\overline{EF}\), then \(\overline{GF}<\overline{EF}\) and \(\overline{EG}+ \overline{GF}=\overline{EF}\)
Interior Angle Axiom
If \(H\) is inside \(\angle EFG\), then \(\angle EFH < \angle EFG\) and \(\angle HFG< \angle EFG\) and \(\angle EFH+ \angle HFG= \angle EFG\).
Pasch’s Axiom
If a line passes between two vertices of a triangle and does not go through the other vertex, then it passes between it and one of the passed vertices.
To be between \(E\) and \(F\) means to be on the segment they define, \(\overline{EF}\), but at neither endpoint. To be inside \(\angle EFG\) means to be between points on \(\overrightarrow{FE}\) and on \(\overrightarrow{FG}\) , with neither point being \(F\). It is instinctive that all humans know what it means for a point to be between two points and – in the case of Pasch’s axiom – also what it means for a segment to be continuous; that is, with no gaps where another segment might slip through. Triangles and quadrilaterals are defined to be convex; the segment between two points interior to two sides is inside the figure. This means that they are not allowed to be concave or degenerate. Interior angles are greater than zero and less than straight, so triangles are never segments and quadrilaterals are never triangles or darts.
In Geometry–Do, between, inside, plane, point, shortest path and straight are undefined terms. These are concepts that a parent does not have to explain to a child; they are just giving names to concepts that are already in the child’s mind. Area is defined as the number of squares that fill a triangle or union of triangles. Like the ancient Greeks, we do not have a rigorous definition of limits but just rely on intuition; wheat plants are infinitesimal compared to fields, so weighing the wheat is almost like calculating a limit. Thus, area too is something that small children can understand without explanation. Defining area as the product of a rectangle’s sides waits for Volume Three: Geometry with Multiplication. This definition of area is not intuitive to small children, who know nothing of multiplication, which is why it is introduced late in our study.
Degrees of angle will not be defined anywhere in Geometry–Do because doing so is trigonometry.
Triangle Inequality Theorem
Any side of a triangle is shorter than the sum of the other two sides.
In ancient Greece, Epicurus scoffed at Euclid for proving a theorem that is evident even to an ass (donkey), who knows what the shortest path to a bale of hay is. Indeed, it is a direct result of our definition that a segment is all the points along the shortest path between two points. It is an exercise for yellow belts to prove it using the greater angle and greater side theorems, but we will satisfy both Epicurus and Euclid by introducing it among the axioms while calling it a theorem.
The foundations explained above are sufficient through bluebelt study. In these early chapters, students will learn to bisect, trisect and quadrisect a segment, and to multiply it by small natural numbers by using repeated addition. No more of these repeated additions are needed than four, for construction of the Egyptian or 3–4–5 right triangle, the only exception being that we mention in passing the 5–12–13 right triangle, which is used by plumbers when installing 22.5° elbows.
Beginners, especially construction workers anxious to complete whitebelt geometry, are advised not to get too hung up on these foundations, which are a bit abstract. But it is essential that we lay a solid foundation for our science. It is recommended that students read again about foundations when they are orange belts and are more comfortable with abstract reasoning. (Also, SSS and ASS, mentioned in the first paragraph, will then be known to them.) By then, those who are not – the construction workers – will be gone. Red belts are expected to teach beginning students to relieve black belts of this task. Pedagogical instruction is provided to red belts for this purpose, and they are also asked to read this foundational material yet again, and deeply.
Black belts will learn of similarity and prove the triangle similarity theorem, which Common Core students take as a postulate because they do not really know how to prove anything. Similarity opens up a whole new world in geometry! Specifically, black belts will go beyond bisecting and trisecting segments to constructing segments whose length relative to a given unit is any rational number. But, for this, another axiom is needed. We have said that a set with both an equivalence relation, \(=\) , and a total ordering, \(\leq\) , is called a magnitude. But to construct segments whose length relative to a given unit is any rational number, length must also be Archimedean.
Archimedes’ Axiom
Given any two segments \(\overline{EF}\) and \(\overline{GH}\), there exists a natural number,\(n\) , such that \(n\overline{EF}>\overline{GH}\).
This may seem trivially true, but Galois (finite) fields are not Archimedean. Every school boy is taught that Archimedes claimed that, given a long enough lever and a fulcrum to rest it on, he could move the world. They typically receive no clear answer from their teacher on why it matters, since no such fulcrum exists and Archimedes seems to ignore that gravity is attractive. The point that Archimedes is making is that, if there were such a fulcrum and much gravity under it, he would need a lever \(6\times10^{22}\) longer on his side of the fulcrum to balance his mass against the Earth. If the fulcrum were one meter from Earth, Archimedes would be in the Andromeda galaxy if he stood on the other end of that long lever. \(6\times10^{22}\) is a big number, but it does exist.
We said above that undefined terms are concepts that one does not have to explain to a child; the adult is just giving names to concepts that are already in the child’s mind. But defining natural numbers as \(1,2,3..\) is only intuitive up to as many fingers as the child has. When I took my fouryearold to another town, she was surprised that a different man was driving the bus. She thought that the few dozen people she had met in our town represented everybody in the world; that is, she thought that the natural numbers are a Galois field modulo \(47\). We think \(6\times10^{22}\) exists because countably infinite fields are consistent; but so are big Galois fields. This axiom is why it is traditional in America to tell children that every snowflake is unique^{9}. That Archimedes’ axiom is not intuitive to small children is one reason why similarity is delayed until black belt.
\(\rho, \sigma\) 
\( \rho\) is right and \(\sigma\) is straight. They are pronounced rho (row) and sigma. 
\(E, F, G, .., Z\) 
Points. Do not use A, B, C, D, R; they have assigned meanings. 
\(M\) 
\(M\) is usually a midpoint; subscripts denote of what; e.g. M_{EF} is of \(\overline{EF}\). 
\(\omega\) 
A circle, pronounced omega. \(O\) is usually a circle’s center. 
\(\bot, \parallel,∦\) 
Perpendicular, parallel and not parallel 
White Belt Instruction: Foundations
Side–Angle–Side (SAS) Theorem
Given two sides and the angle between them, a triangle is fully defined.
Proof
By the segment postulate, the segments have two endpoints and, since they form an angle, they share an endpoint. This is three so, by the triangle postulate, the triangle is fully defined. Congruence is transitive, so any two anywhere are congruent.
Euclid had five postulates, not six, but proof of his fourth proposition, SAS congruence, relied on superposition, which tacitly assumes a whole slew of additional and unmentioned postulates. Many have cast doubt on Euclid, pointing out that superposition – sliding figures around and flipping them over to position one on top of the other – is nowhere defined.
Robin Hartshorne (p. 2), the most famous American geometer of the 20th century, wrote in 2000, “Upon closer reading, we find that Euclid does not adhere to the strict axiomatic method as closely as one might hope… The method of superposition… cannot be justified from the axioms… we can develop geometry according to modern standards of rigor.” But, when Common Core was formulated, Hartshorne was shunted aside because Bill Gates was offering big money to redefine congruence in terms of transpositions – sliding figures around on a computer screen to superimpose them one on top of the other – assuring that geometry ceases to exist the moment a student rises from his school computer. Ask such a student for a formal proof that the bridge in Lake Havasu City, Arizona, USA is congruent to the one that spanned the Thames River in 1831!
Isosceles Triangle Theorem
If two sides of a triangle are equal, then their opposite angles are equal.
Proof
Given \(\overline{EFG}\) with \(\overline{GE}\) and \(\overline{GF}\) equal, by the line postulate, draw \(\overrightarrow{GE}\) and \(\overrightarrow{GF}\) and cut them off at \(H\) and \(I\) so \(\overline{EH}=\overline{FI}\) and \(\overline{GH}=\overline{GI}\) by addition. By SAS,\(\overline{FGH}\cong\overline{EGI}\). By congruence, \(\angle GHF=\angle GIE\) and \(\overline{HF}=\overline{IE}\). By SAS,\(\overline{EHF}\cong\overline{FIE}\) and \(\angle FEH=\angle EFI\). By subtraction, if the exterior angles are equal, then so are the interior angles,\(\angle FEG=\angle EFG\). .
Observe that, when we cite SAS, the triangle vertices are ordered by the side, angle and side that are equal; e.g. \(\overline{FGH}\cong\overline{EGI}\) by SAS means that \(\overline{FG}=\overline{EG}\) and \(\angle FGH=\angle EGI\) and \(\overline{GH}=\overline{GI}\) are given and other equalities, such as \(\angle GHF=\angle GIE\) and \(\overline{HF}=\overline{IE}\), are implied. These implications should be clear without saying, “corresponding magnitudes of congruent triangles are equal.”
The isosceles triangle theorem has a converse, but we wait to become yellow belts before proving it. It is rarely cited while the isosceles triangle theorem is a geometer’s bread and butter.
Proof of the SSS theorem will use a proof by contradiction; that is, show that Y not true X and true is contradictory. We have defined dichotomy and trichotomy; now we assume that G and H are distinct and then consider the four places where H can be if it is not G. Like aiming a rifle at a target, there are only five possibilities: a bull’s eye or a miss to the left, right, above or below. We show that the latter four are impossible. The lemma is based on what “inside” means.
Equilateral Triangle Theorem
If three sides of a triangle are equal, then all three angles are equal.
Proof of the SSS theorem will use a proof by contradiction; that is, show that Y not true and X true is contradictory. We have defined dichotomy and trichotomy; now we assume that G and H are distinct and then consider the four places where H can be if it is not G. Like aiming a rifle at a target, there are only five possibilities: a bull’s eye or a miss to the left, right, above or below. We show that the latter four are impossible. The lemma is based on what “inside” means.
Lemma 1.1
If a triangle is inside another triangle, it has less area.
Side–Side–Side (SSS) Theorem
Given the three sides, a triangle is fully defined.
Proof
Given \(\overline{EFG}\) and \(\overline{EFH}\) with \(\overline{EG}=\overline{EH}\) and \(\overline{FG}=\overline{FH}\), suppose that \(G\) and \(H\) are distinct. By lemma 1.1, if \(H\) is inside \(\overline{EFG}\) or inside the angle vertical to \(\angle EGF\), then \(\overline{EFH}<\overline{EFG}\) or \(\overline{EFH}>\overline{EFG}\), which implies \(\overline{EFG}\neq\overline{EFH}\). Suppose \(H\) is on the \(E\) side of \(\overline{FG}\) but not inside \(\overline{EFG}\).\(\overline{EG}=\overline{EH}\), so \(\overline{EGH}\) is isosceles. \(\angle EHG=\angle EGH\) by the isosceles triangle theorem. By analogous reasoning, \(\overline{FGH}\) is isosceles and thus \(\angle FGH=\angle FHG\).
\(\angle EHG=\angle FHG+\angle EHF\) 
and by analogous reasoning 
\(\angle FGH=\angle EGH+\angle FGE\) 
\(\angle EHG>\angle FHG\) 

\(\angle FGH>\angle EGH\) 
\(\angle EHG>\angle FHG\) 

\(\angle FGH>\angle EHG\) 
A contradiction; \(H\) on the \(F\) side of \(\overrightarrow{EF}\) but not inside \(\overline{EFG}\) is also contradictory.
Fully defined does not imply existence – the longest segment must be less than the sum of the other two – but it does imply uniqueness so, by transitivity, any two anywhere are congruent.
In the following construction, existence and uniqueness of \(\overrightarrow{FE}\) and \(\overrightarrow{FG}\) requires invoking the line postulate, though this goes unsaid. In the same way that, given \(E\) and \(F\), we speak of \(\overline{EF}\) without bothering to invoke the segment postulate, we now speak of \(\overrightarrow{FE}\) or \(\overleftrightarrow{FE}\) whenever \(\overline{FE}\) has been defined. This practice is in keeping with our plan to avoid tedious proofs with mincing steps, but the student should never forget that Euclid’s postulates are everpresent and needed.
Construction 1.1
Bisect an angle.
Solution
Given \(\angle EFG\), take any point H on \(\overrightarrow{FE}\). There exists a point I on \(\overrightarrow{FG}\) such that \(\overline{FH}=\overline{FI}\). Construct an isosceles triangle with base \(\overline{HI}\) and \(J\) apex on the other side of \(\overline{HI}\) from \(F\). By SSS, \(\overline{HFJ}\cong\overline{IFJ}\), which holds the equality \(\angle HFJ=\angle IFJ\)
To construct an isosceles triangle when the base is given, a geometer sets his compass to any length longer than half the base and draws arcs from each endpoint. Where these arcs intersect is an apex; there are two possible, one on each side of the base. These arcs are each called a locus, and together, loci. To construct an isosceles triangle when the apex angle is given, lay off the same arbitrary length on both of the rays from the vertex and then connect these points.
Construction 1.2
Bisect a segment.
Solution
Given \(\overline{EF}\), construct an isosceles triangle with \(\overline{EF}\) as base. Using C. 1.1, bisect the apex angle. Since the sides are equal, there is no need to find H and I; just use E and F. By SAS, the angle bisector intersects \(\overline{EF}\) at its midpoint.
Construction 1.3
Raise a perpendicular from a point on a line.
Solution
Given \(\overleftrightarrow{EF}\) with G on it, lay off the same arbitrary length to the left and to the right of G and then construct an isosceles triangle with this base. By SSS, the segment connecting the apex with G forms congruent triangles and the angles at G are right.
Construction 1.4
Drop a perpendicular from a point to a line.
Solution
Given G not on \(\overleftrightarrow{EF}\), construct an isosceles triangle with apex G and base collinear with \(\overleftrightarrow{EF}\). Using C. 1.1, bisect the apex angle. By SAS it forms congruent triangles and the angles at its intersection with \(\overleftrightarrow{EF}\) are right.
These constructions are the four basic techniques that will be used in combination throughout geometry. At the most fundamental level, all four are much alike. This is analogous to how the jab, hook, uppercut and cross are the basic techniques that are used in combination throughout boxing. But all four involve giving somebody a poke in the nose, so they are much alike.
The perpendicular bisector of a segment is called its mediator. The perpendicular from a triangle vertex to the (extension of the) opposite side is called an altitude. Red belts will extend altitudes past the opposite side and will extend the bisector of an interior angle past the opposite side, but when lengths are assigned to an altitude or to an angle bisector, it means the length of the segment from the vertex to the opposite side. The symbol for perpendicular is .
Center Line Theorem
An angle bisector and a perpendicular bisector coincide if and only if the triangle is isosceles.
Proof
Assume the angle bisector and perpendicular bisector coincide. By SAS (reflexivity, the right angle postulate and bisection), the two right triangles are congruent and so their hypotenuses are equal. Thus, the given triangle is isosceles.
Assume the triangle is isosceles. By the isosceles triangle theorem, the base angles are equal. Construct a median from the apex. By SAS(opposite sides, opposite angles and bisection), the two triangles are congruent. The apex angle is bisected and the angles at the foot of the median are equal; right, because they bisect a straight angle.
The center line is the mediator of the base and the apex angle bisector of an isosceles triangle.
Construction 1.5
Replicate an angle.
Solution
Construct an isosceles triangle with the given angle as its apex angle by laying off equal lengths and connecting them. By SSS, reconstruct this triangle elsewhere.
Construction 1.6
Given a ray and a point on the angle bisector, find the other ray of the angle.
Solution
Draw an arc through the point, then a circle around it through the intersection of the arc and the ray, then a ray through the other intersection of the arc and the circle.
Interior and Exterior Angles Theorem
The bisectors of an interior and exterior angle of a triangle are perpendicular to each other.
Proof
Given \(\overline{EFG}\) and \(H\) on \(\overrightarrow{EF}\) past \(F\),\( \angle EFG\) is the interior angle and \(\angle HFG\) is the exterior angle at vertex \(F\). By C. 1.1, find \(I\) and \(J\) on the angle bisectors of \(\angle EFG\) and \(\angle HFG\), respectively. \(\angle EFI=\angle GFI\) and \(\angle EFJ= \angle HFJ\), so \(\angle EFI+\angle EFJ=\angle GFI+\angle HFJ\) by addition. The union of these four angles is a straight angle and, if a straight angle is cut in two equal angles, then each one is right; thus, \(\angle EFI+\angle EFJ=\rho\) and \(\overrightarrow{FI}⊥\overrightarrow{FJ}\).
Reflection Theorem
If \(\overline{EFG}\) is isosceles with base \(\overline{EF}\), then \(\overline{EFG}\cong\overline{FEG} \) and \(\overline{EFG}\sim\overline{FEG} \)
Proof is left as an exercise. Orange belts will study similarity in more depth and black belts will learn that similar triangles, defined by their angles being equal, imply that their sides are proportional. This is an incredible achievement because angles and lengths are different magnitudes. There is nothing in our postulates to hint that these very different magnitudes might, through hard work and perseverance, eventually be proven to have some relation to each other. For now, just know that congruence implies similarity, but not the converse.
Problem 1.1
Draw a line through a point so it makes equal angles with the sides of an angle.
Solution
By the Isosceles Triangle Theorem Converse, the desired line is the base of an isosceles triangle with the given angle at its apex. By the Center Line Theorem, the base is perpendicular to the apex angle bisector.
Problem 1.2
A fink truss consists of an equilateral triangle built on the middle third of the ceiling joists. The rafters rest on the walls and meet at the triangle apex. Beams from the feet of the triangle meet the rafters at right angles. Draw it. The boards need not have width.
The fink truss leaves the middle open for storage or for a mattress, making it the best choice for a shed or a mountain cabin. For flatter roofs, drop a vertical from the apex and from each rafter midpoint and angled beams from the center to the rafter midpoints. The strongest is the king post truss, a vertical from a right apex and angled beams from its foot to the rafter midpoints.
Problem 1.3
Suppose your girlfriend asks you to install a wall mirror. She is six feet tall in heels and her eyes are six inches below the top of her hair. What is the smallest mirror that allows her to see her entire self and how high should it be above the floor? Does it matter how far away she stands?
Figure this problem out before you turn the page! Don’t just looking at the illustration!
Construct two isosceles triangles with bases from her eyes to her feet and to the top of her hair.
Prove the following theorem. It is a biconditional and so it requires two independent proofs, as did the center line theorem. The student must assume that a point is on the perpendicular bisector and prove that it is equidistant from the endpoints of the segment; and, in a separate proof, he must assume that a point is equidistant from the endpoints of a segment and prove that it is on the perpendicular bisector. The two proofs may be done in either order.
Technically, p and q are equivalent even if proof that q implies p requires citing the previously proven statement that p implies q. However, students see it as a trick if I say, “prove that p and q are equivalent,” but I do not mention that they must prove that p implies q first, and then prove that q implies p. No tricks! If this is the case, then I will call the statement that p implies q a theorem, and the statement that q implies p its converse, but I will not call them equivalent.
But, today, the two statements really are equivalent; prove them in either order. Good luck!
Mediator Theorem
A point is on the perpendicular bisector iff it is equidistant from the endpoints of the segment.
Also, the last paragraphs of White Belt Geometry for Construction Workers demand a practical application of geometry that a framer might encounter on a job site. Students with only an academic interest in geometry, including those who hope to one day compete in the International Mathematical Olympiad (IMO), should not disdain bluecollar work. If you cannot make a wall square and vertical, then you will never win the IMO! Anyway, you need work in the summer.
Construct each triangle using only the information given about it.
1.4 Construct a right triangle given the lengths of the legs.
1.5 Construct a triangle given the lengths of the three sides.
1.6 Construct a triangle given the apex angle and the lengths of the legs.
1.7 Construct a triangle given the lengths of the base, the median to the base and one leg.
1.8 Draw a king post roof truss with a right apex. The boards need not have width.
1.9 Ancient hieroglyphics describe a 350’ tall pyramid that no longer exists. Could a Common Core student prove it congruent to the Luxor hotel in Las Vegas by using transposition?
Comparison with Common Core Geometry
Common Core teachers present the isosceles triangle theorem after showing students the button on Geometer’s Sketchpad for bisecting a segment. They never demonstrate bisecting a segment with compass and straight edge; they rely heavily on that magical midpoint button.
Isosceles Triangle Theorem
If two sides of a triangle are equal, then their opposite angles are equal.
Common Core Proof
Given \(\overline{EFG}\) with \(\overline{GE}\) and \(\overline{GF}\) equal, use C. 1.1 to bisect \(\overline{EF}\) at H. By SSS, \(\overline{HEG}\cong\overline{HFG}\) and thus \(\angle HEG=\angle HFG\)
Alternate Proof
Given \(\overline{EFG}\) with \(\overline{GE}\) and \(\overline{GF}\) equal, by SSS, \(\overline{EFG}\cong\overline{FEG}\) and thus \(\angle EFG=\angle FEG\)
The alternate proof is easier but, for reasons known only to David Coleman, was not chosen for Common Core. Both require SSS and thus neither can be used in Geometry–Do because the proof of SSS requires the isosceles triangle theorem. Coleman dodges the charge of circular reasoning by the simple expedient of not proving SSS. It, SAS and AAS (needed for the converse) are all just factoids for the students to memorize. Cheater! Cheater! Booger eater!
Isosceles Triangle Theorem Converse
If two angles of a triangle are equal, then the opposite sides are equal.
Common Core Proof
We are given \(\overline{EFG}\) with \(\angle E\) and \(\angle F\) equal. Use C. 1.1 to bisect \(\angle FGE\) and extend it to H on \(\overline{EF}\). By AAS, \(\overline{HEG}\cong\overline{HFG}\) and thus \(\overline{EG}=\overline{FG}\).
Common Core states the triangle similarity theorem as an axiom – we prove it in the black belt chapter – calling it either the similarity axiom or the dilation axiom, and then state without proof the AA, SAS and SSS similarity theorems. SAS, SSS, AAS and HL (they do not state ASA because it is not needed for the isosceles triangle theorem and its converse, which is the last theorem they intend to prove) are then just special cases of the similarity/dilation axiom with the scale (dilation factor) being the multiplicative identity – which requires assuming the field axioms for real numbers – and the midsegment theorem is a special case with the scale (dilation factor) being one half. Common Core students who claim to know of easier proofs to the isosceles triangle theorem and its converse can only say this because they did not have to prove SAS, SSS and AAS.
The orangebelt chapter concludes with a section on how to pass a standardized exam of the type that is designed for Common Core students. Most of the people now reading these lines will not survive orange belt, so I will here tell you how a Geometry–Do white belt can pass Common Core exams. First, recognize that it is really an algebra exam in disguise, so review Algebra I. But the big secret is to bring a centerfinding metric ruler and a compass to the exam so you can construct the figures – the ones provided are purposefully wrong – and measure the unknown quantity.
Varsity Tutors Advanced Geometry Exam, problem #14, is solved below, first using geometry and then using the algebra that masquerades as geometry in Common Core. Which is easier?
Problem 1.10
Given a triangle with base units and legs and units, what is the height?
Geometry Solution
Use SSS to construct the triangle and then measure its height. It’s 12!
Algebra Solution
Let x and y be the projections of the 13 and 15 unit legs on the base, respectively. Then \(x+y=14\) and, by the Pythagorean theorem, \(132=x^{2}+h^{2}\) and \(152=y^{2}+h^{2}\). Solve both equations for \(h^{2}\), set them equal and substitute \(y=14x\) into the latter equation.
\(169X^{2}=225(14x)^{2}\)
\(=225196+28xX^{2}\)
\(0=140+28x \)
\(x=\frac{140}{28}=5\)
Substitute \(x=5\) into the first equation, \(132=x^{2}+h^{2}\), and then solve it for h.
\(h=\sqrt{16925}=\sqrt{144}=12\)
Varsity Tutors considers this advanced because almost no American geometry student can answer it correctly or, if they do, it takes them thirty minutes to work through all the algebra. But, if you construct the geometric figure with a ruler and compass (Duh! It’s a geometry exam!), you can solve it in one minute using the most basic whitebelt theorem you know.
Teachers! If you have read this far hoping for advice on how to get your #%$^@ students through the Common Core standardized exam, here it is: Ask for the perimeter of a triangle with vertices \((2,3), (4,4), (7,1)\) and make it a race. The easy way is to lay the three sides endtoend on a line. Taking the sum of three applications of the algebraic distance formula is the hard way.
\(\sqrt{(2(7)^{2})+(3(1)^{2})}+\sqrt{(2(4)^{2})+(3(4)^{2})}+\sqrt{(7(4)^{2})+(1(4)^{2})}\approx18\)
FirstDay Exam in Geometry
The first task of the American highschool geometry teacher is to disabuse students of the notion that geometry is just a boring review of Algebra I. (Nothing new here. Blah!!!) Ask for the perimeter of a triangle with vertices \((2, 3), (4,4), (7,1)\) and make it a race.
The easy way is to lay the three sides endtoend on a line. Put the compass pin at \((7,1)\) and rotate it to lay off the lower left side on the horizontal. Without moving the pin, measure the upper left side and lay it off on the horizontal past the one you just did. Finally, measure the upper right side and lay it off on the horizontal past the one you just did. It’s segment addition!
\(\sqrt{(2(7)^{2})+(3(1)^{2})}+\sqrt{(2(4)^{2})+(3(4)^{2})}+\sqrt{(7(4)^{2})+(1(4)^{2})}\approx18\)
\(\sqrt{(5)^{2}+(4)^{2}}+\sqrt{(2)^{2}+(7)^{2}}+\sqrt{(3)^{2}+(3)^{2}}\)
\(\sqrt{25+16}+ \sqrt{4+49} + \sqrt{9+9}\)
\(\sqrt{41}+\sqrt{53}+ \sqrt{18}\)
\(\approx 6.40 + 7.28 + 4.24\)
\(\approx 17.92\)
White Belt Geometry for Construction Workers
George Birkhoff’s axioms are called metric because they assume the field axioms for real numbers; those of David Hilbert and myself are called intrinsic because they do not. Birkhoff is assuming tape measures longer than one’s work space that do not droop and protractors that measure angles to such precision that they can be projected across one’s work space and the opposite side of the triangle is as accurate as can be measured with one’s tape. Such protractors do not exist. Such tapes exist for cabinet makers, but not for construction workers. The socalled Egyptian triangle can verify that an angle is right, but it does not create a right angle. Finding the corners of a rectangle can be frustrating for carpenters who know only this. It works only if the sides are rigid and reach across the entire work space, so there is no extrapolation error. The only time I recommend that construction workers use the Egyptian triangle is if they build an 8’ wall, nail it to the floor, measure 6’ from it, and then have two men stretch a tape diagonally; when their tape measures 10’, nail the wall to the ceiling joists. It is vertical!
Problem 1.11 Make a 16’ square cabin exactly square.
Solution
To make the cabin face south, stand at the SW corner and aim 45° minus magnetic declination off magnetic north; e.g. in Los Angeles, aim for 33° east. Lay off the diagonal, 22’ 7.5”, in this orientation. Have a man at each stake hold the end of a tape measure while you stretch them out to 16’ and pound stakes in where they intersect.
Squaring a 16’ cabin is easy, but the general rectangle requires greenbelt geometry. Since no construction worker has ever survived the greenbelt entrance exam, I will here break my vow against using unproven theorems and just present a cookbook recipe. A chalked string can be extended six times longer than a tape measure and, because the string is light weight, it does not droop when stretched across these long distances. Because a general rectangle may be several times longer than your tape measure, you will need two strings in addition to your tape. If the strings are new and consistently taunt, you can achieve accuracy to one part in a thousand.
Squaring a foundation must be achieved with no auxiliary lines outside it. This is because it may be in a hole if it is for a basement, or it may be surrounded by trees or cliffs if a plot of land was cleared and graded for a house being built in a forest or cut into a hillside. Parallelism will be defined later, but I will here describe orienting the front of the house with a compass. To make the house face a road, give the front the same compass heading as the center line of the road. To make the house face south, stand at the SW corner and aim 90° minus magnetic declination off magnetic north; e.g. in Los Angeles, aim for 78° east.
Problem 1.12
Square a house’s foundation before pouring the concrete floor.
Solution
Mark the front segment, \(\overline{EF}\), with two stakes measured with a tape and oriented with a compass to be parallel to a road or to the eastwest line; do not neglect declination. Loop the end of string S_{1} over the Estake, stretch it across the front and tie it to the Fstake. Drive a stake, O , into the ground near the center, but slightly towards the front and slightly towards the stake. Loop the end of string S_{2} over the Ostake, stretch it to the Fstake, pinch it with your fingers and then swing this radius around the Ostake until the arc intersects \(\overline{EF}\). Celebrate the construction of an isosceles triangle by driving in another stake at this intersection, E_{1}. Do not lose your pinchedoff length! Lift the S_{1} string off the Estake and loop it over the E_{1}stake. Stretch it over and past the center stake, O; simultaneously, swing string S_{2} around the ostake to point in the opposite direction, away from E_{1}. Stretch both strings so they coincide (lie on top of each other) and drive a stake,G_{1}, in at the end of the length pinchedoff on S_{1}. \(\angle E1F1G1\) is right. Loop the end of string S_{1} over the Fstake, stretch it into ray \(\overrightarrow{FG1}\) and drive a stake G on this ray past G_{1} to where a tape measures the length of the side of the house. Pinch off this length, \(\overline{FG}\) , lift the S_{1} string off the Estake and loop it over the Fstake. Lift the S_{2} string off the Ostake, loop it over the Fstake, pinch off the length \(\overline{EF}\), then lift it off the Fstake and loop it over the Gstake. Stretch both strings out and where their pinched off lengths intersect, drive a stake, H . \(\overline{EFGH}\) is a rectangle of the house’s dimensions.
This may seem complicated, but it leads directly to a rectangle while the Pythagorean theorem converse (if ,\(u^{2}+v^{2}=w^{2}\) then the triangle with these sides is right) is hit and miss.
Many come to Geometry–Do with prejudice against deductive logic. Now is the time to rid ourselves of these losers! They are baggage we will not need to bring to yellowbelt geometry.
Put construction workers and others who come to geometry with an open mind on Team Euclid. Put those who have closed their minds to deductive logic and believe only in coordinate geometry on Team Prástaro. In two classrooms, push the desks to the corners, staple butcher paper to the ceiling and draw a chalk line on it. Give each team a yardstick, two spools of chalked string and two ladders. A team that can draw a chalk line on the floor directly underneath the one on the ceiling gets an A, else an F. Test their answers with a plumb bob.
If the losers on Team Prástaro demand a tape measure instead of a yard stick, explain that, unless you are building an outdoor toilet, rulers are always less than the length of one’s work space.
The Egyptian or 3–4–5 Right Triangle
In the preceding section I wrote, “Finding the corners of a rectangle can be frustrating for carpenters who know only this.” So true! I remember when I was eight that my father had my mother, my brother and I at stakes marking three corners of the foundation of the basement for our house. He kept measuring sides one at a time with his only tape measure and ordering a stake moved a few inches this way or that. The Pythagorean theorem never came out exact and it offered no hints on how to move the stakes to make it exact. Bad day!
In Volume Three: Geometry with Multiplication, the Pythagorean theorem will be expressed as \(u^{2}+v^{2}=w^{2}\) with \(u,v,w\) being real numbers. However, real numbers were only introduced in the 1800s and the modern theory of rational numbers did not precede them by much. Yet Egyptologists assure us that triangles with sides of 3,4 and 5 units appear in fourthousandyearold hieroglyphics. We will do the ancient proof and, in Volume Three, we will do it rigorously.
Egyptian Triangle Theorem
A triangle with sides three, four and five times a unit length is right.
Proof
By C. 1.3, raise a perpendicular from the endpoint of a segment. Quadrisect the segment and lay off three of these units on the perpendicular. Connect the endpoints and lay off these units on it. Observe that it takes exactly five units to fill it.
An analogous proof shows that a triangle of sides 5,12 and 13 is right; this was unknown to the Egyptians. Plumbers can use this triangle when installing 22.5° elbows. Integer solutions to the Pythagorean theorem are known as Pythagorean triples. Students should be aware that Euclid devised a formula that generates Pythagorean triples:\(u=m^{2}n^{2}\),\(v=2mn\),\(u=m^{2}+n^{2}\) for positive integers \(m>n\). Verification is basic algebra; that \(ku,kv,kw\) for k=1,2.. gets them all is advanced. Try it with n=1 and m even, or n=2 and m odd. Know that Fermat’s Last Theorem states that there are no integer solutions to \(u^{n}+v^{n}=w^{n}\) for integers n>2.
Never use a,b,c for the sides of right triangles! The next step after applying the Pythagorean theorem is to use the quadratic formula \(ax^{2}+bx+c=0\) to solve . Giving a,b,c two meanings is the principal cause of confusion. While leaning on a shovel at construction sites, I have seen “master” carpenters almost come to blows because they messed up a right triangle problem.
3–4–5 right triangles are ubiquitous in Common Core. Varsity Tutors Advanced Geometry Exam, problem #22 gives a rhombus of sides 5 inscribed in a rectangle with height 4 and asks the area.
Basic Principles for Design of Wood and Steel Structures
As a geometer, you may be asked to design structures like gates, towers, gantries or bridges.
Wrong! 
Everybody knows that a diagonal is required to make a rigid triangle, but a drive through the country indicates that few know which way it goes. Wood beams can withstand a tremendous compressive load – 1700 psi for Douglas Fir – but cannot lift a load because the screws pull out. Wire rope is just the opposite. \(\frac{1}{8}\) ” can lift 340 pounds but flexes when compressed. Similarly, metal tubes bend under any large compressive load and, once kinked, they are useless. 
Wooden diagonals go from the foot of the gate post upwards and wire rope diagonals go from the top of the gate post downwards. For a wooden tower to be rigid it must have crossed wooden diagonals so there are some that are angled upwards towards any direction of wind.
A gantry is two Aframes with a beam between them and a hoist that slides along the beam. Mimicking the allsteel commercial ones with wood does not work because, when overloaded, the legs spread apart and bow outwards. The crossbar is pulling them together, which is not what wood does well. Instead, run wire rope from each foot to the midpoint of the other leg.
A drive through the country indicates that almost all wooden gates have collapsed. This is because they have a wooden diagonal angled downwards and it reaches across the entire 12’ or 14’ gate, making too horizontal an angle. Also, failed gates were overengineered on the latch side, adding unnecessary weight far from the hinges. 1” planks are all it takes to stop cattle.
The gate shown on the next page is 14’ wide for farm equipment and is designed to stop cattle, not people. The wire rope loops through the eyes and around both sides of the gate. Solid lines are 2”thick boards or 4”thick posts, dashed lines are 1”thick planks. Note that the boards and posts are all assembled edgewise, so their widest sides are coplanar.
The C boards are inset into B and glued with wooden dowels to add strength. There are five hinges and the gate post is rectangular; two hinges attached to a round post are weak. In the winter the ground freezes to the frost line and, in the spring, the top few inches thaw but do not drain through the frozen ground below, which is why it is so muddy. Water that soaks into the gate post can only drain out the bottom if it extends below the frost line. Also, there should be gravel, not concrete, below it to aid drainage. Gates often collapse because the post rots.
For automotive bridges too high to be supported with pillars, put a 4” x 6” x 12’ post vertical and two 4” x 4” posts at 45 under the center of each of the two stringers and lift them with 0.5” steel cables attached to eye bolts in the concrete footers. The two vertical posts should have crossed braces – it is a mistake to look only at the side view and neglect twisting forces.Yellow belts will learn to build stone bridges cut from river rocks that are strong enough to support truck traffic!
Detailed plans for wooden foot bridges of various sizes are available. Sorry Grasshopper, but, while the plans are free, the hula girl coming out to dance on your completed bridge is extra.
Fortification Construction as an Application of Geometry
Steel cannons with rifled bores brought an end to statesponsored castle construction, but some of what was learned during the time of smoothbore bronze cannons is still relevant today for people engaged in lowintensity conflicts^{10}. By lowintensity I mean, by mutual consent, both the home owners and the bandits restrict themselves to small arms, usually defined as 7.62 mm rifles and hand grenades, because they are under a real army, but it will ignore smallarms fire.
The first principal of building fortifications is to build them for yourself, not for the enemy. If you dig trenches or set out some Jersey barriers, they may stop the enemy’s wheeled vehicles, but they will also provide cover for enemy infantry. Thus, you should have a vertical retaining wall facing rearward and an earthen glacis slope facing forward. Flatten the top of a gently rounded hill, digging deep enough that the cutdown area requires a three to fourfoot high retaining wall all around it. The windows are high enough that the defenders can graze the slope with rifle fire, but the attackers cannot fire at the base of the house wall until they crest the retaining wall.
The second principal is to not have blind spots. The defenders should have bastions protruding from the corners of the building so attackers cannot press themselves up against the wall and be hidden from the windows. But, if the bastions are round, like the turrets in a medieval castle, there are blind spots directly in front of them. They should be tapered, like the points on a star. 

The third principal is that stone shatters when hit by bullets, but concrete and brick do not. Also, landscape with crushed stone to make walking noisy. Get rid of boulders that can be thrown.
There is little application for geometry in the design of fortifications, but whitebelt geometers should be familiar with the basics. Green belts will learn of machine gun emplacement, which really does require geometry. It would be embarrassing for the Geometry–Do practitioner to boast of these advanced techniques while showing ignorance of basics like glacis slopes.
Yellow Belt Instruction: Congruence
Angle–Side–Angle (ASA) Theorem
Given two angles and the included side, a triangle is fully defined.
Proof
Given \(\overline{EFG}\) and \(\overline{HIJ}\) with \(\angle GEF=\angle JHI\), \(\overline{EF}=\overline{HI}\) and \(\angle EFG=\angle HIJ\), let us assume that \(\overline{GE} and \overline{JH}\) are not equal. Suppose that \(\overline{GE}>\overline{JH}\). By the interior angle axiom, there is a point K between G and E such that \(\overline{KE}=\overline{JH}\). \(\overline{EFK}\cong\overline{HIJ}\) by SAS and, by congruence, \(\angle EFK=\angle HIJ\). But \(\angle EFK<\angle EFG\) because K is inside \(\angle EFG\), which is given to be equal to \(\angle HIJ\); a contradiction. Suppose that \(\overline{GE}<\overline{JH}\). Proof that this is impossible is the same but has K between J and H. Thus, \(\overline{GE}=\overline{JH}\) and, by SAS, \(\overline{EFG}\cong\overline{HIJ}\). .
It is an easy corollary that isosceles triangles have two angle bisectors equal. Prove it, please.
Isosceles Triangle Theorem Converse
If two angles of a triangle are equal, then the opposite sides are equal.
Proof
Given \(\overline{EFG}\) with \(\angle E=\angle F\), \(\overline{EFG}\cong\overline{FEG}\) by ASA and thus \(\overline{GE}=\overline{GF}\)
Orange belts study parallel lines, which are everywhere equidistant. For this statement to mean something, we must define the distance from a point to a line. There are many points on a line that define segments to a point not on the line. Which segment? The one of minimum length!
Perpendicular Length Theorem
The perpendicular is unique and is the shortest segment from a point to a line.
Part One
There is only one segment from a point to a line that is perpendicular to it.
Proof
Drop a perpendicular from G to \(\overleftrightarrow{EF}\). Label the intersection H and extend the perpendicular that much again to I. Let J be on \(\overleftrightarrow{EF}\) and not H. By SAS, Drop a perpendicular from G to \(\overleftrightarrow{EF}\). Label the intersection H and extend the perpendicular that much again to I. Let J be on \(\overleftrightarrow{EF}\) and not H. By SAS, \(\overline{GHJ}\cong\overline{IHJ}\) and, by congruence, \(\angle GJH=\angle IJH\). By the segment postulate, \(\angle GJI\) is not a straight angle because \(\angle GHI\) is straight and it is axiomatic that two points fully define a segment. If \(\angle GJI\) is not straight, then \(\angle GJH\), which is half of \(\angle GJI\), is not right.
Dropping a perpendicular uses C. 1.1, extending a segment uses the line segment postulate and adding an equal segment uses addition. Thus, SAS means that \(\overline{GH}=\overline{IH}\), \(\angle GHJ=\angle IHJ\) by the right angle postulate, and \(\overline{HJ}=\overline{HJ}\) by reflexivity. All of this seems obvious and is omitted for the sake of brevity. I am avoiding the tedious and condescending style of highschool textbooks. .
Part Two
The perpendicular is the shortest segment from a point to a line.
Proof
\(\overline{GI}<\overline{GJ}+\overline{JI}\) by definition of segment. Halve both sides so \(\overline{GH}<\overline{GJ}\)
Vertical angles are across from each other at an intersection; this is not the physics definition.
Vertical Angles Theorem
Given \(\overleftrightarrow{EF}\) and \(G\), \(H\) on opposite sides of it, \(G\),\(E\),\(H\) are collinear iff a pair of vertical angles is equal.
Proof
Assume G, E, H are collinear and I is on \(\overleftrightarrow{EF}\) with E between it and F. Then, \(\angle GEF\) and \(\angle GEI\) are supplementary. \(\angle GEH\) is straight; thus, \(\angle GEF\) and \(\angle HEF\) are supplementary. \(\angle GEI\) and \(\angle HEF\) are both supplementary to \(\angle GEF\) and thus equal.
Assume I is on \(\overleftrightarrow{EF}\) with E between it and F; also, \(\angle GEI=\angle HEF\). I, E, F are collinear and thus \(\angle GEI\) and \(\angle GEF\) are supplementary. By substitution, \(\angle HEF\) and \(\angle GEF\) are supplementary; hence, G, E, H are collinear.
Exterior Angle Inequality Theorem
The exterior angle of a triangle is greater than either remote interior angle.
Given \(\overline{EFG}\), extend \(\overline{EF}\) to H. We claim that \(\angle EGF < \angle HFG\) and \(\angle GEF < \angle\) HFG.
Part One
\(\angle FGE < \angle HFG\)
Proof
Bisect \(\overline{FG}\) at I and extend the median past I to double its length at J. Connect F and J. \(\angle JFI < \angle HFI\) because \(\angle HFJ+\angle JFI=\angle HFI\). By the vertical angles theorem and SAS, \(\overline{GIE}\cong\overline{FIJ}\), which holds the equality \(\angle EGI=\angle JFI\). By substitution, \(\angle EGI<\angle HFI\). By substitution of points collinear on the side of an angle, \(\angle EGF<\angle HFG\)
Proof of part two is left as an exercise for the student. Just bisect \(\overline{FE}\) instead of \(\overline{FG}\). It is an easy corollary that the base angles of an isosceles triangle are half the exterior to the apex angle.
The exterior angle inequality theorem is a foundational theorem that is needed to prove AAS congruence and the transversal theorem and cyclic quadrilateral theorem converse.
Greater Angle Theorem
If two sides of a triangle are unequal, then their opposite angles are unequal, the shorter side opposite the smaller angle and the longer side opposite the larger angle.
Proof
Given \(\overline{FG}<\overline{EF}\) in triangle \(\overline{EFG}\), find H between E and F such that \(\overline{HF}=\overline{FG}\) and connect it to G. By the Exterior Angle Inequality Theorem, \(\overline{FEG}<\overline{FHG}\). By the isosceles triangle theorem, \(\angle FHG=\angle FGH\). Also, \(\angle FGH<\angle FGE\) because the former is inside the latter. Thus, \(\angle FEG<\angle FHG=\angle FGH<\angle FGE\). Simplifying, \(\angle FEG<\angle FGE\).
Greater Side Theorem
If two angles of a triangle are unequal, then their opposite sides are unequal, the smaller angle opposite the shorter side and the larger angle opposite the longer side.
The word angle or side in the name refers to the result, not the given information. Proof of the greater side theorem is left as an exercise; it is analogous to that of the greater angle theorem.
Triangle Inequality Theorem
Any side of a triangle is shorter than the sum of the other two sides.
As discussed in the Note to Philosophers, this is a direct result of the definition of a segment. It can also be proven using the greater angle and greater side theorems, as can the hinge theorem. A corollary is that diameters are the greatest chords. (They try not to let it go to their heads.)
Hinge Theorem
Given two triangles with two corresponding sides equal, the included angle in one is smaller than in the other if and only if the opposite side is shorter in the former than in the latter.
Some geometers cite the triangle inequality theorem where I say “by definition of segments.” It is easier to define a segment as the shortest path between two points than to draw triangles among noncollinear points. Problems 3.3 and 3.4 are examples of this method. Shortest path has to be intuitive because, until Volume Three, we cannot associate real numbers with lengths.
Angle–Angle–Side (AAS) Theorem
Given two angles and a side opposite one of them, a triangle is fully defined.
Proof
Given \(\overline{EFG}\) and \(\overline{HIJ}\) with \(\angle E=\angle H, \angle F=\angle I\) and \(\overline{FG}=\overline{IJ}\), assume that \(\overline{HI}\neq\overline{EF}\); specifically, that \(\overline{HI}<\overline{EF}\). Thus, there is a K between E and F such that \(\overline{KF}=\overline{HI}\). By SAS, \(\overline{KFG}\cong\overline{HIJ}\), which holds the equality \(\angle GKF=\angle JHI\). But \(\angle JHI=\angle GEF\) is given, so \(\angle GKF=\angle GEF\). \(\angle GKF\) is an exterior angle of \(\overline{EKG}\) equal to a remote interior angle of \(\overline{EKG}\), which contradicts the Exterior Angle Inequality Theorem. Thus, the assumption that \(\overline{EF}\neq\overline{HI}\) cannot be true. By SAS, \(\overline{EFG}\cong\overline{HIJ}\).
Isosceles Altitudes Theorem
Two altitudes are equal if and only if the triangle is isosceles.
Hypotenuse–Leg (HL) Theorem
Given the hypotenuse and one leg of a right triangle, it is fully defined.
Proof
By C. 1.3, raise a perpendicular from a line the length of the leg and connect its endpoint to the line on both sides with segments the length of the hypotenuse. By the isosceles triangle theorem, the base angles are equal, and by AAS, the triangles are congruent.
ASS cannot be a congruence theorem because it lacks uniqueness. Even if \(\overline{EFG}\cong\overline{HIJ}\) (their areas, sides and angles are all equal), \(\overline{EFG'}\) can be constructed with the same angle, side and side and it is not congruent to \(\overline{HIJ}. \overline{HL}\) initially seems to lack uniqueness because we can construct two triangles, \(\overline{EFG}\) and \(\overline{EFG'}\), with\(\angle E\) right, \(\overline{EF}\) the leg and \(\overline{EG}=\overline{EG'}\) the hypotenuse. But \(\overline{EFG}\cong\overline{EFG'}\) and so we know that \(\overline{EFG}\cong\overline{HIJ}\) iff \(\overline{EFG'}\cong\overline{HIJ}\) because congruence is transitive.
Faced with tremulous homeschool parents who intended to teach geometry using his textbook but were unsure that they could, Kiselev bolstered their courage by casting a magic spell on them:
Those reading these lines are hereby summoned to raise their children to a good command of Elementary Geometry, to be judged by the rigorous standards of the ancient Greek mathematicians.
You will be pleased or perhaps appalled to learn that I too can cast magic spells… actually curses. The Aguilar Curse: Students who try to use the ASS “theorem” will grow donkey ears.
A student who tried to use the ASS “theorem” on an exam. Don’t let this be you!
No worries. We will next prove the ASL theorem, which is almost like the ASS “theorem,” but its use avoids the embarrassment of walking through high school hallways sporting donkey ears.
Lemma 2.1
The sum of any two interior angles of a triangle is less than a straight angle.
Proof
Consider \(\angle E\) and \(\angle F\) in \(\overline{EFG}\) and let H be on the extension of \(\overline{EF}\) so \(\angle HFG\) is exterior to \(\angle EFG\). By the Exterior Angle Inequality Theorem, \(\angle FEG<\angle HFG\). By supplementarity, \(\angle EFG+\angle \overline{HFG}\) is straight and, by substitution, \(\angle EFG+\angle FEG\) is less than straight.
This lemma is absolute geometry in the sense that Bolyai used the term to mean what is common to Euclidean and hyperbolic geometry. But it is not true in elliptic geometry. Some people use the term to include all three geometries. We will just not cite ASL until the next chapter.
Angle–Side–Longer Side (ASL) Theorem
Given an angle and the side opposite the angle not less than a near side, a triangle is fully defined.
Proof
Given \(\overline{EFG}\) and \(\overline{HIJ}\) with \(\angle E=\angle H, \overline{EF}=\overline{HI}\) and \(\overline{FG}=\overline{IJ}\), assume that \(\overline{HJ}\neq\overline{EG}\); specifically, that \(\overline{HJ}<\overline{EG}\). Thus, there is a K between E and G such that \(\overline{EK}=\overline{HJ}\). By SAS, \(\overline{EFK}\cong\overline{HIJ}\), which holds the equality \(\overline{FK}=\overline{IJ}\). But \(\overline{FG}=\overline{IJ}\) is given, so \(\overline{FK}=\overline{IJ}\) by transitivity and \(\angle FKG=\angle FGK\) by the isosceles triangle theorem. By lemma 2.1, their sum is less than a straight angle and, since they are equal, they are acute; hence, \(\angle EKF\) is obtuse by supplementarity. By lemma 2.1, \(\angle EKF\) is larger than both \(\angle KEF\) and \(\angle KFE\) because together they must sum to an acute angle. By the greater side theorem, \(\overline{EF}\) is the longest side, which contradicts the assumption that the side opposite \(\angle E\) is longer than it. Thus, the assumption \(\overline{HJ}\neq\overline{EG}\) is wrong.
Given \(\overline{EFG}\) and \(\overline{HIJ}\) with \(\angle E=\angle H\), \(\overline{EF}=\overline{HI}\) and \(\overline{FG}=\overline{IJ}\), if \(\angle G\) and \(\angle J\) are both obtuse, right or acute, then \(\overline{EFG}\cong\overline{HIJ}\). This is an alternative way to state the ASL theorem. Next is OSS, a corollary of ASL that is cited in the classic proof of the SteinerLehmus theorem. It states that ASS works for obtuse triangles.
Obtuse Angle–Side–Side (OSS) Theorem
Given an obtuse angle and two sides that are not bracketing it, a triangle is fully defined.
Proof citing ASL
By lemma 2.1, the obtuse angle is larger than either of the other angles so, by the greater side theorem, the side opposite it is longer than either other side. Thus, by ASL.
Proof citing HL
By supplementarity, the angle exterior to the given angle is fully defined. Drop a perpendicular from the vertex between the given sides to the extension of the unknown side. It intersects the extension because the given angle is obtuse and forms a triangle outside the given triangle that, by AAS, fully defines the length of the altitude. By HL, the union of the two triangles is fully defined and, by subtraction, the given one is.
We now prove a fundamental theorem that will be cited throughout, and then some theorems about circles that do not cite any version of the parallel postulate. Thus, potential students of nonEuclidean geometry can read the white and yellowbelt chapters without learning of any theorems (except ASL, which contradicts elliptic geometry) that will contradict their later studies.
Angle Bisector Theorem
A point is on an angle bisector if and only if it is equidistant from the sides of the angle.
Proof
Given the angle bisection, the point is equidistant by AAS; the converse by HL.
Diameter and Chord Theorem
A diameter bisects a chord if and only if the diameter is perpendicular to the chord.
Proof
If a diameter bisects the chord, then, by SSS, the two triangles formed by connecting radii to the endpoints of the chord are congruent and, by the mediator theorem, the diameter is perpendicular to the chord. Suppose the diameter is perpendicular to the chord; by HL, the two triangles to either side of it are congruent and the chord is bisected.
Equal Chords Theorem
In the same or equal circles, equal chords are equally distant from the center, and the converse.
Proof
Suppose the chords are equal. Draw radii to the endpoints and drop perpendiculars from the center to the chords. By the diameter and chord theorem, these perpendiculars bisect the chords and thus the half chords are equal. By HL, the half triangles are congruent and so the distances from the chords to the center are equal
Proof of the converse is left as an exercise.
Common Chord Theorem
If two circles have a common chord, its mediator is the line of centers.
Proof
By the diameter and chord theorem, the mediator of the chord is a diameter to both circles and on the line of centers.
The common chord theorem is also true if the chord is reduced in length to a point. If two circles touch at a single point they are tangent; they may be sidebyside or one inside the other.
Tangent Theorem
A line intersects a circle where it is perpendicular to the radius iff that is a point of tangency.
Part One
If a line intersects a circle where it is perpendicular to the radius, then that is the only point where it intersects the circle.
Proof
Label the point where the line intersects the circle and is perpendicular to the radius T. Suppose it intersects at another point, S. Find U on the line such that \(\overline{ST}=\overline{TU}\). Extend the radius across the circle to be a diameter. By the diameter and chord theorem, \(\overline{SU}\) is a chord. Thus, \(S, T, U\) are distinct points on a line and also on a circle, a contradiction because lines and circles are different things and cannot coincide at three points.
Part Two
If a line intersects a circle only once, then it is perpendicular to the radius to that point.
Proof
Because it intersects only once, every other point on the line is outside the circle and thus farther from the center than the radius. By the perpendicular length theorem, only the perpendicular is the shortest segment from a point to a line.
So, we know something about the tangent point, but how do we find it?
Construction 2.1
Through a point outside a circle, draw a line tangent to the circle.
Yellow Belt Solution
Let E be the center of the given circle, F be the given point and G be the intersection of \(\overline{EF}\) with the circle. Draw a concentric circle that passes through \(F\). By C. 1.3, raise a perpendicular from \(G\) and call its intersection with the larger circle \(H\). Let \(I\) be the intersection of \(\overline{EH}\) and the given circle; \(I\) is the desired tangent point.
Proof
By SAS, \(\overline{GEH}\cong\overline{IEF}\), which holds the equality \(\angle G=\angle I. \angle G\) is right by construction, so \(\angle I\) is right and, by the tangent theorem, \(I\) is a tangent point.
When the point is far from the circle (e.g. a machine gun and its kill circle), the outer circle may go over the edge of the paper. Also, one’s compass may not be large enough to draw this big circle. The greenbelt solution requires an arc only half this radius and extending no farther than the given circle, so it will be the standard technique. But, even if yellow belts do not have the most efficient method, they have a method, so everything we describe in this chapter is feasible.
Common Point Theorem
An intersection of two circles is a tangent point if and only if it is on the line of centers.
Proof
If it is a tangent point, then, by the tangent theorem, it is perpendicular to the radii from both centers and, by supplementarity, it is on the line of centers. If it is on the line of centers, then, by the perpendicular length theorem, the tangent there is unique.
This theorem is like the common chord theorem but for circles that just touch. It is needed by construction workers who wish to blend arcs smoothly together; e.g. for concrete curbs that partition flower beds. An asphalt company might use this theorem to pave an Sshaped driveway to a detached garage. Orange Belt Geometry for Construction Workers uses this theorem to prove that the haunch and the crown arcs of a Tudor or generic arch blend smoothly together.
Two Tangents Theorem
Two tangents from an external point are equal and their angle bisector intersects the center.
Proof
By the tangent theorem, the points of tangency are where the radii are perpendicular to the tangents and so, by the perpendicular length theorem, the center is equidistant from the tangents. By the angle bisector theorem , their angle bisector intersects the center and, by HL, the two triangles are congruent. Hence, the tangents are equal.
Tangent Bisection Theorem I
If two circles touch, the perpendicular to the line of centers through the circles’ touching point cuts their common tangents in half.
Blue belts will prove tangent bisection theorem II about the extension of the common chord.
It is an easy corollary of ASA that isosceles triangles have two angle bisectors equal; while not among Euclid’s propositions, it must surely have been known to him. Is the converse also this easy? No. Geometers waited over 2000 years for the converse, which is known as the SteinerLehmus theorem. Here is not the classic proof of Jakob Steiner, but a better one because it does not invoke the parallel postulate. It is made possible by Pasch’s axiom, which was introduced by Moritz Pasch in 1882, about forty years after Lehmus proposed and Steiner solved the problem.
Chords Inside Circles Theorem
Chords are inside their circle.
Proof
Let \(H\) be inside chord \(\overline{EF}\), center \(O\). \(\angle OEH=\angle OFH\) by the isosceles triangle theorem. \(\angle OHF>\angle OEH\) by the greater angle theorem, so \(\angle OHF>\angle OFH\) and, by the greater side theorem, \(\overline{OF}>\overline{OH}\); that is, \(H\) is closer to the center than a point on the circle.
This may seem obvious, but it is needed to prove that Roman and Gothic arches exist.
Mirror Problem
Given a light and an observer, find the point on a mirror to shine the light at the observer.
Solution
Let \(L\) be the light and \(O\) be the observer. Drop a perpendicular from \(O\) to the mirror or its extension and extend it an equal distance to \(O'\). This point is called the reflection of \(O\). Let \(P\) and \(M\) be the intersections of \(\overline{LO}'\) and \(\overline{OO}'\) with the mirror, respectively, and \(M'\) a point on the mirror past \(P\) from \(M\). By SAS, \(\overline{OMP}\cong\overline{O'MP}\), so \(\angle OPM=\angle O'PM\). by the Vertical Angles Theorem \(\angle O'PM=\angle LPM'\). Thus, \(\angle OPM=\angle LPM'\).
Note that, in geometry, a reflection is a point defined by another point and a line; the observer is the reflection of the point behind the mirror that the light is aimed at, and viceversa. People usually use the term reflection to refer to the light that the observer sees, but his position and the virtual point behind the mirror are reflections of each other. If a laser is shined at an arc, it reflects in the same way as it would off a flat mirror that is tangent to the arc at the aiming point.
Problem 2.1
We wish to pound an anchor between two poles with guy wires to the tops of each pole to reinforce them. Where should we position the anchor to use the shortest possible wire?
Solution
Guess at where the anchor should be and draw in the guy wires. Draw a guy wire to one pole’s reflection. By SAS the reflected triangles are congruent and so their hypotenuses are equal. By definition of segment, if the anchor is not collinear with the end of one pole and the end of the other pole’s reflection, it is badly guessed. Correct it.
The guy wire is a physical representation of a light beam. Thus, the mirror property is that reflections preserve distance, so the light beam in the mirror problem travels the same distance if it bounces off the mirror to the observer or if it penetrates and goes to the observer’s reflection.
Problem 2.2
From a house in the country, construct a dirt road to a straight paved road, the latter twice as fast as the former, to minimize travel time to a nearby town on the paved road.
Solution
Guess at where the intersection should be. On the other side of the paved road, construct a half equilateral triangle with the hypotenuse on the paved road to town and the short leg extending into the dirt. From the intersection, you could get to the right angle or to town in the same time. By definition of segment, if the intersection is not collinear with the house and the right triangle, it is badly guessed. Correct it.
This is an important lesson, Grasshopper. If asked to minimize the sum of segments that are not collinear, use congruence theorems to show that this is equal to the sum of segments that can be made collinear. By the definition of segment, the shortest path has all straight joints. We guessed at the answer, drew a figure, saw what was wrong with the guess and then corrected it.
Problem 2.3
Given \(\angle EFG\) and \(H\) within it, find points on each ray such that the perimeter of the triangle they make with \(H\) is minimal.
Solution
Drop perpendiculars from \(H\) onto \(\overrightarrow{FE}\) and onto \(\overrightarrow{FG}\), then lay off that much again past \(\overrightarrow{FE}\) and \(\overrightarrow{FG}\) to \(H'\) and \(H''\), respectively. Let \(I=\overline{H'H''}∩\overline{FE}\) and \(J=\overline{H'H''}∩\overline{FE}\). \(\overline{HIJ}\) has perimeter \(\overline{H'H''}\), which is of minimal length.
Constructing a triangle so its perimeter is minimal will become the bread and butter of the Geometry–Do practitioner; he should be able to just look at a problem like P. 2.3 and immediately construct the solution. But the thought process of guessing at the solution and then seeing what is wrong with the picture so it can be redrawn correctly should not be forgotten. (Here, if we had guessed at \(I\) and \(J\), then drawn \(\overleftrightarrow{IJ}\), laid off \(\overline{HI}\) past \(I\) to \(H'\) and laid off \(\overline{HJ}\) past \(J\) to \(H''\), our “solution” would violate the perpendicular length theorem.) If the student becomes too blasé about the initially easy problems, he will later encounter a difficult problem and feel like he has run into a wall. But, if he thinks carefully on the easy problems and treats each one as training for more difficult problems that he knows are coming, he will later discover himself doing problems that other geometers consider difficult without having ever noticed his own passage.
Minimal Base Theorem
Given the apex angle and the sum of the legs, the triangle with minimal base is isosceles.
Solution
Given \(\overline{EFG}\) with \(\overline{GE}=\overline{GF}\), half the given length, lay off \(x\) on \(\overrightarrow{GE}\) and subtract it from \(\overrightarrow{GF}\) to points \(H\) and \(I\), respectively. \(\overline{GE}+\overline{GF}=overline{GH}+\overline{GI}\). Drop perpendiculars from \(H\) and \(I\) to \(\overleftrightarrow{EF}\) with feet \(H'\) and \(I'\), respectively. By the Isosceles Triangle Theorem Converse and the Vertical Angles Theorem, \(\angle HEH'=\angle IFI'\). By AAS, \(\overline{H'EH}\cong\overline{I'FI}\), so \(\overline{EF}=\overline{H'I'}\). But \(\overline{H'I'}<\overline{HI}\) by the triangle inequality theorem. Thus, \(\overline{EF}\) is the shortest possible base.
Problem 2.4
Given the center, draw a circle that bisects a given circle so their common chord is its diameter.
Solution
Draw the line of centers and, by C. 1.3, a perpendicular through the center of the given circle. By the common chord theorem, this diameter is the circles’ common chord. Draw the circle from the given center with a radius to the endpoints of the chord.
Problem 2.5
Through one of the two points of intersection of two equal circles, draw two equal chords, one in each circle, forming a given angle.
Solution
Bisect the given angle and replicate this on both sides of the common chord. The sides of this angle extended make two chords, one in each circle, forming the given angle.
Proof
Let \(O_{1}, O_{2}\) be the centers, \(\overline{FG}\) their common chord, M its midpoint and \(F_{1}, F_{2}\) the endpoints of the constructed chords \(\overline{FF_{1}}, \overline{FF_{2}}\) in circles \(O_{1},O_{2}\), respectively. By the Common Chord Theorem and by HL, \(\overline{O_{1}FM}\cong\overline{O_{2}FM}\), and so \(\angle O_{1}FM=\angle O_{2}FM\). Thus, \(\angle F_{1}FO_{1}= \overline{O_{1}FM±O_{1}FF_{1}}\) and \(\angle F_{2}FO_{2}= \overline{O_{2}FM} ±\overline{O_{2}FF_{2}}\), and so \(\angle F_{2}FO_{1}=\angle F_{2}FO_{2}\). By the diameter and chord theorem, \(\angle O_{1}M_{F_{1}F}F\) and\( \angle O_{2}M_{F_{2}F}F\) are right. By AAS, \(\overline{M_{F_{1}F}FO_{1}} \cong \overline{M_{F_{2}F}FO_{2}}\), so \(\overline{M_{F_{1}F}F}=\overline{M_{F_{2}F}F}\), and so \(\overline{FF_{1}}=\overline{FF_{2}}\)
If the angle is narrow and goes between the centers, then the proof is easier because it can cite the vertical angles theorem. But we must take every possibility into consideration!
Common Core textbooks are fond of the acronym CPCTC, which stands for Corresponding Parts of Congruent Triangles are Congruent. I do not use this acronym for two reasons:
1. It is wrong. It should be Corresponding Magnitudes of Congruent Triangles are Equal, or CMCTE. “Parts” is nowhere defined; they are magnitudes. Using the term congruent for both triangles and magnitudes to the complete exclusion of the term equal is wrong. Congruence and equality are different things; there is good cause for having two words.
2. It is excessively verbose. When I say, “by SAS, \(\overline{EFG}≅\overline{HIJ}\) and so \(\overline{EG}=\overline{HI}\),” the students know immediately from how the vertices are ordered that \(\overline{EF}=\overline{HI}\) and \(\angle EFG=\angle HIJ\) and \(\overline{FG}=\overline{IH}\). Sometimes I will say, “\(\overline{EFG}≅\overline{HIJ}\), which holds the equality \(\overline{IH}EG=\overline{HI}\).” Students know that we say that two triangles are congruent because congruence implies that corresponding magnitudes are equal; they do not need to be repetitively told this.
Common Core textbook authors purposefully make proofs as boring and tedious as possible – displaying a fetishlike attention to reflexivity – because they have already decided that, in the remaining 95% of their textbook, they are going to turn into demagogues and just announce theorems without proving them. In How Math Can Be Taught Better I mock a Common Core textbook that takes five steps to prove that two angles, both given as right, are equal. Duh!
Let us now illustrate another method for solving geometry problems: Solve the problem with algebra and then replicate the algebra with geometry.
Construction 2.2
Construct two segments given their sum and their difference.
Solution
Add the sum and the difference. Bisect this to get the longer segment and then cut off the difference to get the shorter segment.
\((x+y)+(xy)=2x\) Thus, we add the sum and the difference and then bisect it to get the longer side. \(x(xy)=y\) Thus, we cut off the difference to get the shorter segment.
For the most part I despise Common Core, but one practice that I will adopt is their insistence that math classes include some classic literature. Who hasn’t read The Pit and the Pendulum?
I could no longer doubt the doom prepared for me by monkish ingenuity in torture… [The pendulum’s] nether extremity was formed of a crescent of glittering steel, about a foot in length from horn to horn; the horns upward, and the under edge evidently as keen as that of a razor.  Edgar Allen Poe
Problem 2.6
If the horns of Poe’s pendulum are at points E and F one moment and then at points E' and F' a minute later, where is the axle from which the pendulum is suspended?
Solution
By the diameter and chord theorem, the mediators to chords \(\overline{EE'}\) and \(\overline{FF'}\) are diameters. Their intersection is the common center of the two concentric circles.
Note that \(\overline{EF}\) and \(\overline{E'F'}\) are not chords; there is no assurance that the horns rise equally above the blade. The endpoints of each chord are the same vertex of the figure at different times. Each point on the pendulum makes its own circle around an axle common to them all.
It is an easy corollary of ASA that isosceles triangles have two angle bisectors equal; while not among Euclid’s propositions, it must surely have been known to him. Is the converse also this easy? No. Geometers waited over 2000 years for the converse, which is known as the SteinerLehmus theorem. Here is not the classic proof of Jakob Steiner, but a better one because it does not invoke the parallel postulate. It is made possible by Pasch’s axiom, which was introduced by Moritz Pasch in 1882, about forty years after Lehmus proposed and Steiner solved the problem.
Steiner–Lehmus Theorem
If a triangle has two angle bisectors equal, then it is isosceles.
Modern Proof
Given \(\overline{EFG}\) with the angle bisectors \(\overline{FH}=\overline{EI}\), assume that \(\overline{GE}\neq\overline{GF}\); specifically, that \(\overline{GE}<\overline{GF}\). By the greater angle theorem, \(\angle F<\angle E\), and their halves,\(\angle HFG<\angle IEG\). By C. 1.5, replicate \(\angle HFG\) with one ray \(\overrightarrow{EI}\) and the other inside \(\angle IEG\); let J be its intersection with \(\overline{GF}\). From \(\angle HFG<\angle IEG\) and insideness, \(\angle EFJ<\angle FEJ\). By the greater Side Theorem, \(\overline{EJ}<\overline{FJ}\), so one can lay off \(\overline{FK}=\overline{EJ}\) with K inside \(\overline{FJ}\). By SAS, \(\overline{JEI}\cong\overline{KFH}\), which holds the equality \(\angle EJI=\angle FKH\). Applying Pasch’s Axiom to the line \(\overleftrightarrow{EJ}\) and the triangle \(\overline{KGH}\), there exists a point L that is on \(\overleftrightarrow{EJ}\) and between H and K. In the triangle \(\overline{LJK}\), the interior angle \(\angle LJK\) equals the exterior angle \(\angle LKF\) because \(\angle EJI=\angle FKH\). This contradicts the exterior angle theorem and so \(\overline{GE}=\overline{GF}\).
Difficult? It is easier than the classic proof that has auxiliary lines outside the triangle and cites greenbelt theorems. It is made possible because we accept “inside” and “between” as intuitive.
For the pedagogic purpose of teaching older students how to teach younger students, and because of its importance to the history of geometry, the classic proof is given at the beginning of the redbelt chapter. This is of interest to students who intend to take a historyofmath class.
1. From a house in the country, construct a dirt road (30 mph limit) to a straight paved road (50 mph limit) to minimize travel time to a nearby town on the paved road.
2. Given two sides, an angle opposite one of them and the fact that the angle opposite the other side is acute or that it is obtuse, prove that the triangle is fully defined.
3. Solve the following problem in miniature golf. If every bounce is true, this means that the ball ricochets from the wall at the same angle that it strikes the wall.
Construct each figure and prove that it is fully defined.
2.7 Construct a right triangle given a leg and the hypotenuse.
2.8 Construct a right triangle given the hypotenuse and one acute angle.
2.9 Construct a right triangle given the sum and the difference of the legs.
2.10 Construct a triangle given the apex altitude and the legs.
2.11 Construct a triangle given the base and the two base angles.
2.12 Construct a triangle given the base, a base angle and the median to that leg.
2.13 Construct a triangle given the base, the median to the base and one leg.
2.14 Construct a triangle given the apex angle, its angle bisector and a leg.
2.15 Construct two equal circles whose common chord equals their given radius.
2.16 Construct two equal circles given their common chord and the distance between centers
2.17 Let \(\omega1\) and \(\omega2\) be concentric circles with \(\omega1\) inside \(\omega2\) and center O. If E, F are on \(\omega1\) and G, H are on \(\omega2\) and \(\angle OEG=\angle OFH\), prove that \(\overline{OEG}\cong\overline{OFH}\).
2.19 Let \(\overline{EFG}\) be isosceles with \(\overline{EG}=\overline{FG}\) and H and I be the feet of the altitudes to \(\overline{EG}\) and \(\overline{FG}\), respectively. Prove that the ray through their intersection bisects \(\angle G\).
2.19 Let \(\overline{EFG}\) be isosceles with \(\overline{EG}=\overline{FG}\) and let H be a point in the interior of \(\overline{EFG}\) such that \(\angle EHG=\angle FHG\). Prove that H is on the median to \(\overline{EF}\).
Orange Belt Geometry for Construction Workers
Suppose that, while a house is being built, the workers must cross a ditch that is bridged with a \(2” X 12” X 12’\) board. It is bouncy underfoot and dangerously close to breaking when a fat man with a wheelbarrow full of sand is at its center. Inscribe an arc in the board from one corner to the center point of the other side and across to the other corner. Cut along the arc, steam bend another board over it and attach the two boards with corner braces. The radius of an arc inscribed in a \(2” X 12” X 12’\) board is \(19’ 3”\). Do not attempt to draw this arc by flexing a length of PVC pipe because it bends mostly in the center and does not describe an arc.
To inscribe an arc in a rectangle, find the center using the circumcenter theorem with your three points being two corners on the same side and the midpoint of the opposite side. For aesthetic reasons, it is important that the several arcs in your design all have the same radius. So, if one arc is given, choose any three widely spaced points on it and apply the circumcenter theorem.
Arches with one center like this are called Roman and their chord is called the spring line. A Gothic arch has two centers; if they are at the endpoints of the spring line, it is called equilateral. If the centers are on the extensions of the spring line, it is called lancet; if inside, it is called deep.
If you know the height and width, then draw mediators of the legs of the triangle defined by the end points of the spring line and the height of the arc above its midpoint. Where the mediators intersect the spring line or its extension are the two arc centers. If they intersect the spring line before they intersect each other, the centers must be below the spring line equally distant past where the mediators intersect each other. How far past is indeterminate. This is called a pointed Roman arch because it does not look very Gothic.
The Goths were a long time ago and use of their arches gives a building a medieval look. The ogee arch is Gothic. It is very pointed, which is inefficient as a window, but symbolic in a church.
Ogee Arch Circumscribe a rectangle around an equilateral triangle. Draw arcs centered at the midpoint of the base and the two upper corners with a radius of half the base.
In America, a Roman arch is often seen above a window, door or gate, probably because it is the simplest arch and the only one familiar to most construction companies. It is squat, which is fine if it is on top of a rectangle but, if it is the entire opening, then its corners are too sharp. A Tudor arch is also squat, but it has upright corners and works well for wide entrances to big buildings.
Tudor Arch Quadrisect the spring line so \(E, F, G, H, I\) are collinear in that order. Construct an equilateral triangle \(\overline{FJH}\) with \(J\) below the spring line and extend \(\overleftrightarrow{FJ}\) to \(K\) so \(\overline{JK}=\overline{FJ}\) and \(\overleftrightarrow{HJ}\) to \(L\) so \(\overline{JL}=\overline{HJ}\). Draw arcs (called haunch arcs) centered at \(F\) and \(H\) with radii \(\overline{FE}\) and \(HI\) to intersect \(\overleftrightarrow{FJ}\) and \(\overleftrightarrow{HJ}\) at \(M\) and \(N\), respectively. Draw arcs (called crown arcs) centered at \(K\) and \(L\) with radii \(\overline{KM}\) and \(\overline{LN}\) to intersect at the apex. The height of the apex of the Tudor arch is \(\frac{\sqrt{6}\sqrt{3}}{2}\approx35.87%\) of its width, the length of the spring line.
Suppose that a beam bridge made of reinforced concrete spans a canal four meters wide and it is \(h\) meters above the concrete sides of the canal. This is ugly, so the city has hired you, a mason, to construct a façade to make it appear that the bridge is a Tudor arch made entirely of brick.
There are many possible arches with four centers but, if we are going to call ours Tudor, then it must coincide with the classic one when given a height \(\frac{\sqrt{6}\sqrt{3}}{2}\) of the width. Also, we must quadrisect the spring line regardless of the height. If the centers of the haunch arcs are near the edges of the spring line, the arch would look Gothic but with rounded corners. If the centers of the haunch arcs are near the midpoint of the spring line, the arch would look like a pointed Roman. But putting the centers of the haunch arcs on the quartile points makes it look Tudor.
Construction 2.2
Construct a Tudor arch given a height and width approximately that of the classic Tudor arch.
Solution
Quadrisect the spring line so \(E, F, G, H, I\) are collinear in that order. Raise a perpendicular to \(\overline{EI}\) from \(G\) and lay off \(\overline{GK}\) to be the given height. Raise a perpendicular to \(\overline{EI}\) from I and lay off \(\overline{IJ}=\overline{IH}\). Connect \overline{KJ}. Raise a perpendicular to \(\overline{KJ}\) from K and, in the direction of \(\overline{EI}\), lay off \(\overline{KL}=\overline{IJ}\). Connect \overline{HL}. Where its mediator intersects \(\overleftrightarrow{KL}\) is the center of the crown arc, O, and its radius is \(OK\). The haunch arc is centered at H and has radius \(HI\); the arcs meet on \(\overleftrightarrow{OH}\).
Proof that this coincides with the Tudor arch when given a height \(\frac{\sqrt{6}\sqrt{3}}{2}\) of the width comes later.
Suppose that the Tudor arch we just designed was not a façade but was meant to support the roadbed. Would it work? No. Brick is not as strong as stone and – far more fatal to the design – the mortar between them is much weaker than the bricks. Since all the bricks are rectangular, they can only describe an arc if there is mortar between them and it is in a wedge shape. Brick arches look elegant and are recommended for aesthetic reasons, but they are not weight bearing.
The Romans cut stones into isosceles triangle frustums (isosceles triangles cut by lines parallel to their bases) for bridge construction. This works in areas free of earthquakes but, because gravity presses them together, if the bridge is jolted, it all comes crashing down. In modern times, bolting the stones together would make it earthquake resistant, but a more fundamental flaw is that it is only strong if it is most of a semicircle, which is too steep for vehicles. Flatter Roman arches are weak in the center and they push outwards as well as down, so they must have sturdy foundations. Also, if the river rises, flood water pushes on the sides of the bridge near the banks and, if the river is flowing fast, it can push the stones out of place and cause a collapse.
Engineers wish for an arch flat enough for vehicles to climb over but with upright corners so the weight presses straight down and the roadbed is above high waters. An ellipse would work, but this is not feasible because every stone would be unique and it is not practical to readjust the saw for each cut. However, if you have two saws, they can be calibrated to cut the triangle frustum needed for arcs of two different radii. Thus, the Tudors approximated an ellipse with haunch arcs of one radius and a single crown arc that goes all the way across rather than meeting in a pointed apex. With modern construction, Tudor bridges are strong enough for truck traffic.
Tudor Bridge Quadrisect the spring line so \(E, F, G, H, I\) are collinear in that order. Construct an equilateral triangle \(\overline{FJH}\) with J below the spring line and extend \(\overleftrightarrow{FJ}\) and \(\overleftrightarrow{HJ}\). Draw arcs (called haunch arcs) centered at F and H with radii \(\overline{FE}\) and \(\overline{HI}\) to intersect \(\overleftrightarrow{FJ}\) and \(\overleftrightarrow{HJ}\) at M and N, respectively. Draw an arc (called the crown arc) centered at J with radius \(\overline{JM}=\overline{JN}\).
The Tudor bridge is just like the Tudor arch except, instead of extending the legs of the equilateral triangle an equal distance to find the two centers of the crown arcs, we take the apex of the equilateral triangle to be the center of the single crown arc. An isosceles triangle taller/shorter than an equilateral triangle makes the bridge flatter/steeper, but this is not recommended. The height of the classic Tudor bridge is \(\frac{3\sqrt{3}}{4}\approx31.70%\) of its width, the length of the spring line.
Is this design useful to carpenters? No. Archimedes, a Syracuse mathematician who studied in Alexandria shortly after the time of Euclid, invented what is now called the Archimedes’ trammel. It draws ellipses or, if a router is attached to the arm, cuts them. Proof that this works is beyond the scope of this book, but it does, so carpenters have no need for approximating an ellipse with arcs. Note that the word trammel, without the adjective, refers to a board with two awls clamped to it. Pushing one into wood and rotating the board allows the other to scratch an arc, so the device functions just like a geometer’s compass, but it can reach across a sheet of plywood. Geometers drawing postersize figures can buy a pencil trammel at a shop for wood workers.
Construction 2.2 requires that the height and width are approximately that of the classic Tudor arch; that is, in the range \(14w < h < 22w\). To draw an arch to an arbitrary height requires that the radius of the haunch arc be a function of height; specifically, \(\frac{2h}{3}\). This is called generic because squat/tall generic arches look like Roman/Gothic arches with rounded corners and, for the same height as the classic Tudor arch, the generic arch looks Tudorish but it is not exactly the same.
wo thirds of \(\frac{\sqrt{6}\sqrt{3}}{2}\) is \(\frac{\sqrt{6}\sqrt{3}}{2}\approx35.87%≠25%\) of the width, so it is not mathematically Tudor.
We will cite C. 3.8; construction workers must skip ahead and just take it as a cookbook recipe.
Generic Arc Bisect the spring line so \(E, G, I\) are collinear in that order and raise a perpendicular from G to K so that \(\overline{GK}\) is the given height. By C. 3.8, trisect \(GK\) and call twothirds the height x. Locate F and H on the spring line so \(EF\) and \(HI\) are x long and raise a perpendicular from I to J also x long; connect \(JK\). Drop a perpendicular \(\overleftrightarrow{KL}\) with \(\overline{KL}\) also x long. Connect \(\overline{HL}\). Where its mediator intersects \(\overleftrightarrow{KL}\) is the center of the crown arc. The haunch arc is centered at H; the arcs meet on the line from the crown arc center through H.


Tudor Arc 
Generic Arc 
Problem 2.14
Prove that, for any \(x < h \), the generic arch’s haunch and crown arcs are tangent.
Altitude 
The perpendicular from a triangle vertex to the opposite side’s extension

Angle^{11} 
Two rays, called the sides, sharing a common endpoint, called the vertex 

\(\angle F\) if there is one angle at F or \(\angle EFG\) for the angle between \(\overleftrightarrow{FE}\), \(\overleftrightarrow{FG}\)
Acute

An angle that is less than a right angle

Apex

The angle opposite the base of a triangle

Base

In a triangle with a base, the angles at either end

Complementary

Two angles that sum to one right angle

Exterior

The angle supplementary to an interior angle

Interior

An angle inside a triangle or quadrilateral at a vertex

Obtuse

An angle that is greater than a right angle

Right

The bisection of a straight angle

Straight

An angle formed by a line with its vertex being a point on that line

Supplementary

Two angles that sum to one straight angle

Vertical

Angles across from each other at an intersection


Arc 
Part of a circle; within equal circles, angles at the center and the arcs they cut off are a transformation of each other.

Area^{12} 
The number of squares(geometry means “Earth measure” so, initially, individual wheat plants) that fill a triangle or union of triangles; A=\(\overline{EFG}\)

Auxiliary 
Lines or arcs not given whose intersection goes beyond analytic

Axiom 
A proposition that is assumed without proof for the sake of studying the consequences that follow from it

Base 
The side of an isosceles triangle bracketed by the equal angles; b or B
The side of a triangle opposite the apex angle or the one that it is built on, or the side of a quadrilateral called the base for reference purposes.

Bi–Conditional 
A statement of the form p if and only if q. It is true if both p and q are true or both p and q are false. p implies q; also, q implies p. Proof of neither implication can cite the other implication. If and only if is abbreviated iff.

Bisect 
To divide a segment or an angle into two equal parts, called halves

Center Line 
The mediator of the base of an isosceles triangle

Chord 
The segment between two points on a circle

Common 
The segment between the intersection points of two circles

Circle 
All the points equidistant from a point, which is called the center

Collinear 
A set of points that are all on the same line

Concentric 
Two or more circles with the same center but different radii

Congruent 
Two triangles whose areas and whose sides and interior angles are equal;
congruence theorems prove that only three are needed to prove the rest.

Contradiction, Proof by 
To prove that statement X implies statement Y, assume that X is true and Y is not true and show that this is impossible.

Converse 
Given the statement that X implies Y, the statement that Y implies X

Convex 
The segment between two points interior to two sides is inside the figure

Diameter 
A chord that crosses the center of a circle; d or D

Dichotomy 
Proof by contradiction when there are two possibilities

Endpoint 
A point at the end of a segment, arc or ray

Equal 
Comparable magnitudes that are not less than nor greater than each other

Equidistant 
Two pairs of points that define two segments of equal length

Equivalence class 
A set of objects that are equal, congruent, similar or parallel

Relation 
A set and a transitive, reflexive and symmetric relation

Equivalent 
Conditions, any two of which are biconditional

Extend 
Construct the line fully defined by a segment; produce is an archaic term

Figure 
A set of points. They may be alone or joined in lines, segments and arcs.


The intersection when one drops a perpendicular from a point to a line

Fully Defined 
A figure with the given characteristics is unique, if it exists

Hypotenuse 
The side of a right triangle opposite the right angle; \(w\)

Lay Off 
To extend a segment or line a given distance from a point and then cut it

Length 
The unique measure of how long a segment is; \(\overline{EF}\)

Legs 
The sides of a triangle other than the base or the hypotenuse

Lemma 
A theorem used for proving other more important theorems

Line 
A segment extended in both directions; denoted \(\overleftrightarrow{EF}\) if \(\overline{EF}\) is the segment

Line of Centers 
The line that passes through the centers of two circles

Locus 
All the points that satisfy a condition; the plural is loci (lōsī)

Magnitude 
A set with both an equivalence relation, = , and a total ordering, \(\leq\)

Median 
A segment from a vertex of a triangle to the midpoint of the opposite side

Mediator 
The perpendicular bisector of a segment.

Midpoint 
The point where a segment is bisected

NonEuclidean 
A postulate set that contains one that contradicts the parallel postulate

Ordering 
TA set and a relation, \(\leq\), that is transitive, reflexive and antisymmetric
Total \(a \leq b\) or \(b \leq a\) for every a,b in the set

Perpendicular 
A line whose intersection with another line makes a right angle

Postulate 
The axioms that are specific to geometry, not to other branches of math

Quadrilateral 
The union of two triangles adjacent on a side such that it is convex; \(\overline{EFGH}\)

Radius 
A segment from the center of a circle to the circle; plural, radii

Ray 
A segment extended in one direction; denoted \(\overrightarrow{EF}\) if \(\overline{EF}\) is the segment

Reflexive Relation 
A relationship that is always true of something when compared to itself

Relation 
A true/false operator on an ordered pair of elements from a given set

Segment 
All the points along the shortest path between two points; \(\overline{EF}\)

Semisum 
Half the sum of two lengths or of two angles

Side 
One of the three segments that form a triangle

Symmetric Relation 
A relation that can be stated of two things in either order

Tangent 
A line that touches a circle at exactly one point; if its length is referred to, this means the segment between the two points that define it.

Theorem 
A statement requiring proof using postulates or already proven theorems

Touch 
To intersect at exactly one point, neither less often nor more often
Touching Point the tangent point; also called the point of tangency

Transitive Relation 
If a relation is true for a and b and for b and c, then it is true for a and c

Triangle 
The segments connecting three points such that the figure is convex; \(\overline{EFG}\)
Egyptian

A triangle with sides 3, 4 and 5 units long

Equilateral

A triangle with all sides equal

Isosceles

A triangle with two sides equal

Obtuse

A triangle with one angle obtuse

Right

A triangle with one angle right


Trichotomy 
Proof by contradiction when there are three possibilities

Undefined Terms 
Concepts that even small children have an intuitive understanding of; between, inside, plane, point, shortest path and straight

Under Defined 
Not enough given information; the solutions are infinite in number

Vertex 
The intersection of two lines, rays, or sides of a triangle or quadrilateral

Vertical 
Vertical angles; but also for construction workers, parallel to gravity
