Survey of GeometrySupplementary Notes on

Elementary Geometry

Paul Yiu

Department of MathematicsFlorida Atlantic University

Summer 2007

Contents

1 The Pythagorean theorem i1.1 The hypotenuse of a right triangle . . . . . . . . . . . . . . . . i1.2 The Pythagorean theorem . . . . . . . . . . . . . . . . . . . . . iii1.3 Integer right triangles . . . . . . . . . . . . . . . . . . . . . . . iii

2 The equilateral triangle v2.1 Construction of equilateral triangle . . . . . . . . . . . . . . . . v2.2 The medians of an equilateral triangle . . . . . . . . . . . . . . . vi2.3 The center of an equilateral triangle . . . . . . . . . . . . . . . . vii2.4 The circumcircle and incircle of an equilateral triangle . . . . . . vii2.5 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . viii2.6 Some interesting examples on equilateral triangles . . . . . . . . x

3 The square xiii3.1 Construction of a square on a segment . . . . . . . . . . . . . . xiii3.2 The diagonals of a square . . . . . . . . . . . . . . . . . . . . . xiv3.3 The center of the square . . . . . . . . . . . . . . . . . . . . . . xv3.4 Some interesting examples on squares . . . . . . . . . . . . . . xv

4 Some basic principles xvii4.1 Angle properties . . . . . . . . . . . . . . . . . . . . . . . . . . xvii

4.1.1 Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . xvii4.1.2 Angle sum of a triangle . . . . . . . . . . . . . . . . . . . xviii4.1.3 Angle properties of a circle . . . . . . . . . . . . . . . . . xviii

4.2 Tests of congruence of triangles . . . . . . . . . . . . . . . . . . xix4.2.1 Construction of a triangle with three given elements . . . xix4.2.2 Congruence tests . . . . . . . . . . . . . . . . . . . . . . xix

iv CONTENTS

5 Circumcircle and incircle xxiii5.1 Circumcircle . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxiii

5.1.1 The perpendicular bisector locus theorem . . . . . . . . . xxiii5.1.2 Construction of circumcircle . . . . . . . . . . . . . . . . xxiv5.1.3 Circumcircle of a right triangle . . . . . . . . . . . . . . . xxiv

5.2 The incircle . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxv5.2.1 The angle bisector locus theorem . . . . . . . . . . . . . xxv5.2.2 Construction of incircle . . . . . . . . . . . . . . . . . . . xxv5.2.3 The incircle of a right triangle . . . . . . . . . . . . . . . xxvi

5.3 Tangents of a circle . . . . . . . . . . . . . . . . . . . . . . . . xxvi5.3.1 Tangent at a point on the circle . . . . . . . . . . . . . . . xxvi5.3.2 The tangents from a point to a circle . . . . . . . . . . . xxvi

5.4 Some interesting examples on the circumcircle and incircle of atriangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxvii

6 Uses of congruence tests xxix6.1 Isosceles triangles . . . . . . . . . . . . . . . . . . . . . . . . . xxix6.2 Chords of a circle . . . . . . . . . . . . . . . . . . . . . . . . . xxx6.3 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxi

6.3.1 Properties of a parallelogram . . . . . . . . . . . . . . . . xxxi6.3.2 Quadrilaterals which are parallelograms . . . . . . . . . . xxxii6.3.3 Special parallelograms . . . . . . . . . . . . . . . . . . . xxxii

6.4 The midpoint theorem and its converse . . . . . . . . . . . . . . xxxiii6.4.1 The midpoint theorem . . . . . . . . . . . . . . . . . . . xxxiii6.4.2 The converse of the midpoint theorem . . . . . . . . . . . xxxiv6.4.3 Why are the three medians concurrent? . . . . . . . . . . xxxiv

6.5 Some interesting examples . . . . . . . . . . . . . . . . . . . . xxxv

7 The regular hexagon, octagon and dodecagon xxxvii7.1 The regular hexagon . . . . . . . . . . . . . . . . . . . . . . . . xxxvii7.2 The regular octagon . . . . . . . . . . . . . . . . . . . . . . . . xxxviii7.3 The regular dodecagon . . . . . . . . . . . . . . . . . . . . . . xxxix

8 Similar triangles xli8.1 Tests for similar triangles . . . . . . . . . . . . . . . . . . . . . xli8.2 The parallel intercepts theorem . . . . . . . . . . . . . . . . . . xliii8.3 The angle bisector theorem . . . . . . . . . . . . . . . . . . . . xliv8.4 Some interesting examples of similar triangles . . . . . . . . . . xlv

CONTENTS v

9 Tangency of circles xlix9.1 External and internal tangency of two circles . . . . . . . . . . . xlix9.2 Three mutually tangent circles . . . . . . . . . . . . . . . . . . . l9.3 Some interesting examples of tangent circles . . . . . . . . . . . l

10 The regular pentagon liii10.1 The golden ratio . . . . . . . . . . . . . . . . . . . . . . . . . . liii10.2 The diagonal-side ratio of a regular pentagon . . . . . . . . . . . liv10.3 Construction of a regular pentagon with a given diagonal . . . . lv10.4 Various constructions of the regular pentagon . . . . . . . . . . . lvi10.5 Some interesting constructions of the golden ratio . . . . . . . . lviii

Chapter 1

The Pythagorean theorem

1.1 The hypotenuse of a right triangle

This chapter is on the famous Pythagorean theorem. We certainly all know thatthis is an important relation on the sides of a right triangle. The side oppositeto the right angle is called the hypotenuseand is the longest among all threesides. The other two are called the legs. If we denote the legs by a and b, and thehypotenuse by c, the Pythagorean theorem says that

a2 + b2 = c2.

A

B

Cb

ac

Why is this relation true?Suppose you have a cardboard whose length and breadth are 3 inches and

4 inches. How long is its diagonal? Your young students have not known thePythagorean theorem yet. Help them calculate this length without throwing thePythagorean theorem to them.

Take two identical copies of the cardboard and cut each one along a diagonal.In this way there are 4 congruent right triangles. Arrange these 4 triangles in thisway:

ii The Pythagorean theorem

4 1

23

In this figure, the 4 right triangles bound a large square outside and their hy-potenuses bound another smaller square inside.

We calculate areas.(1) The outer side square has side length 3 + 4 = 7 units. Its area is 72 = 49

square units.(2) The 4 right triangles make up 2 rectangular cardboards each with area

3× 4 = 12 square units. The 4 triangles have total area 2× 12 = 24 square units.(3) The inside square therefore has area

49 − 24 = 25

square units. Each of its sides has length 5 units.This method indeed applies to any right triangle with given legs.

Square on hypotenuse = square on sum of legs

− 2 × rectangle with given legs as sides.

b a

b

a

a b

b

ac

c c

c

4 1

23

Why is the inside region a square?

1.2 The Pythagorean theorem iii

1.2 The Pythagorean theorem

Rearrange the “puzzle” in §1.1 in another form:

4

1

2

3a

a

b

b

Here, we see the same 4 right triangles inside the same outer square. Insteadof the inside square, we now have two smaller squares, each built on a leg of theright triangle. This gives the famous Pythagorean theorem:

a2 + b2 = c2.

1.3 Integer right triangles

If we take any two numbers p and q and form

a = 2pq, b = p2 − q2, c = p2 + q2, (1.1)

then we havea2 + b2 = c2.

In particular, if p and q are integers, then so are a, b, and c.Sometimes, integer right triangles constructed from formula (1.1) can be re-

duced. For example, with p = 3 and q = 1, we obtain (a, b, c) = (6, 8, 10), whichclearly is simply the right triangle (3, 4, 5) magnified by factor 2. We say that aninteger right triangle (a, b, c) is primitive if a, b, c do not have common divisor(other than 1). Every integer right triangle is a primitive one magnified by an in-teger factor. A right triangle constructed from (1.1) is primitive if we choose theintegers p, q of different parity 1 and without common divisors.

1This means that one of them is even and the other is odd.

iv The Pythagorean theorem

Here are all primitive integer right triangles with p, q < 10:

p q a b c

2 1 4 3 53 2 12 5 134 1 8 15 174 3 24 7 255 2 20 21 295 46 16 57 27 47 68 18 38 58 79 29 49 8

How many distinct integer right triangles are there whose sides are not morethan 100?

Chapter 2

The equilateral triangle

Notations(O) circle with center OO(A) circle with center O, passing through AO(r) circle with center O, radius r

2.1 Construction of equilateral triangle

An equilateral triangle is one whose three sides are equal in length. The very firstproposition of Euclid’s Elements teaches how to construct an equilateral triangleon a given segment.

A B

C

If A and B are the endpoints of the segments, construct the circles A(B) andB(A) to intersect at a point C. Then, triangle ABC is equilateral.

An equilateral triangle is also equiangular. Its three angles are each equal to60◦.

vi The equilateral triangle

2.2 The medians of an equilateral triangle

LetABC be an equilateral triangle, andM the midpoint ofBC. The segmentAMis a medianof the triangle. Since �ABM ≡ �ACM by thetest, this line AM is also(i) the bisector of angle A,(ii) the altitude on the side BC (since AM ⊥ BC), and therefore(iii) the perpendicular bisector of the segment BC.

B C

A

M

This also means that one half of an equilateral triangle is a right triangle withangles 30◦ and 60◦. We call this an 30 − 60 − 90 right triangle.

In a 30− 60− 90 right triangle, the hypotenuse is twice as long as the shorterleg. The two legs are in the ratio

√3 : 1.

2

1

√3

2.3 The center of an equilateral triangle vii

2.3 The center of an equilateral triangle

Consider two medians of the equilateral triangle intersecting at O.

B C

A

M

NO

Triangles AON and BOM are both 30 − 60 − 90 right triangles. Therefore,OA = 2 ×ON and OB = 2 × OM .However, OA+OM and OB +ON have the same length.This means OM = ON , 1

and OM = 13× AM , ON = 1

3× BN . We have shown that any two medians of

an equilateral triangle intersect at a point which divides each of them in the ratio2 : 1. This is the same point O for all three medians, and is called the center ofthe equilateral triangle.

2.4 The circumcircle and incircle of an equilateraltriangle

Since the three medians of an equilateral triangle have the same length, the centerO is equidistant from the vertices. It is the center of the circumcircle of theequilateral triangle. The same center O is also the center of the incircle of theequilateral triangle.

1AM = BN ⇒ OA+OM = OB+ON ⇒ 2×ON+OM = 2×OM+ON ⇒ OM = ON .

viii The equilateral triangle

B C

A

O

B C

A

O

2.5 Trigonometry

For an acute angle θ, the trigonometric ratios sin θ, cos θ and tan θ are defined by

sin θ =a

c, cos θ =

b

c, tan θ =

a

b,

where (a, b, c) are the sides of a right triangle containing an acute angle θ withopposite leg a.

If any one of the three ratios is known, then the other two can be found. Why?

(1) sin2 θ + cos2 θ = 1.

(2) tan θ = sin θcos θ

.

(3) sin(90◦ − θ) = cos θ.

The precise determination of the trigonometric ratios of an angle is in generalbeyond the scope of elementary geometry. There are, however, a few specialangles whose trigonometric ratios can be determined precisely.

2.5 Trigonometry ix

c

b

a

θ

A C

B

θ 30◦ 60◦

sin θ

cos θ

tan θ

x The equilateral triangle

2.6 Some interesting examples on equilateral trian-gles

(1) For an arbitrary point P on the minor arc BC of the circumcircle of an equi-lateral triangle ABC, AP = BP + CP .

B C

A

P

Q

Proof. If Q is the point on AP such that PQ = PC, then triangle CPQ is equi-lateral since . Note that ∠ACQ = ∠BCP .Thus, �ACQ ≡ �BCP by the test. From this, AQ = BP ,and AP = AQ+QP = BP + CP .

(2) Consider a triangle ABC whose angles are all less than 120◦. Constructequilateral triangles BCX , CAY , and ABZ outside the triangle.

Let F be the intersection of the circumcircles of the equilateral triangles. Notethat ∠CFA = ∠AFB = 120◦. It follows that ∠AFB = 120◦, and F lies onthe circumcircle of the equilateral triangle BCX as well. Therefore, ∠XFC =∠XBC = 60◦, and ∠XFC + ∠CFA = 180◦. The three points A, F , X arecollinear. Thus,

AX = AF + FX = AF +BF + CF

by (1) above. Similarly, B, F , Y are collinear, so do C, F , Z, and BY , CZ areeach equal to AF +BF + CF .

Theorem 2.1 (Fermat point). If equilateral triangles BCX , CAY , and ABZare erected on the sides of triangle ABC, then the lines AX , BY , CZ concur ata point F called the Fermat point of triangleABC. If the angles of triangleABCare all less than 120◦ (so that F is an interior point), then the segments AX , BY ,CZ have the same length AF +BF + CF .

2.6 Some interesting examples on equilateral triangles xi

F

Z

X

Y

A

B C

(3∗) Napoleon’ theorem. If equilateral triangles BCX , CAY , and ABZ areerected on the sides of triangle ABC, the centers of these equilateral trianglesform another equilateral triangle.

Z

X

Y

A

B C

xii The equilateral triangle

Chapter 3

The square

3.1 Construction of a square on a segment

Given a segment AB, construct(1) an equilateral triangle ABP ,(2) an equilateral triangle APQ (different from ABP ),(3) the segment PQ and its midpointM ,(4) the ray AM ,(5) the circle A(B) to intersect this ray at D,(6) the circles B(A) and D(A) to intersect at C.Then ABCD is a square.

A B

CD

PQM

This construction gives an oriented square. If you apply it to AB and BA,you will get two different squares on the two sides of AB.

xiv The square

3.2 The diagonals of a square

A diagonal of a square divides it into two congruent right triangles. Each is anisosceles right triangleor a 45 − 45 − 90 triangle. If each side of a square hasunit length, a diagonal has length

√2.

1

1√

2

1

√2

1

θ 45◦

sin θ

cos θ

tan θ

The diagonals of a square are perpendicular to each other. They divide thesquare into 4 isosceles right triangles.

3.3 The center of the square xv

3.3 The center of the square

The intersection of the diagonals is the center of the square. It is the commoncenter of the circumcircle and the incircle of the square.

3.4 Some interesting examples on squares

(1) If ABCbCa and ACBcBa are squares erected externally on the sides AB andAC of a triangle ABC, the triangle ACaBa has the same area as triangle ABC.

Bc

Ba

Ca

Cb

A

B C

(2∗) The perpendicular from A to BaCA passes through the midpoint of BC.Likewise, those from B to CbAb and from C to AcBc pass through the midpointsof CA and AB respectively. These three perpendiculars meet at the centroidG oftriangle ABC.

xvi The square

G

Bc

Ba

Ca

Cb

Ab Ac

A

B C

(3∗) The midpoint of BcCb does not depend on the position of the vertex A. Itis also the same as the center of the square erected on the BC, on the same sideas A.

Bc

Ba

Ca

Cb

A

B C

M

Chapter 4

Some basic principles

4.1 Angle properties

4.1.1 Parallel lines

Consider two parallel lines �1, �2, and a transversal L.

�1

�2

L

α

β

γ δ

(1) The correspondingangles α and β are equal.(2) The alternate angles β and γ are equal.(3) The interior angles β and δ are supplementary, i.e., β + δ = 180◦.The converses of these statements are also true. In other words, if any one of

(1), (2), (3) holds, then the lines �1 and �2 are parallel.

xviii Some basic principles

4.1.2 Angle sum of a triangle

The angle sum of a triangle is always 180◦. This is because when a side of thetriangle is extended, the external angle formed is equal to the sum of the tworemote internal angles.

A

B Cβ

α

4.1.3 Angle properties of a circle

(1) Let P be a point on the major arc AB of a circle (O).

∠AOB = 2∠APB.

P

A B

O

Q

P

A B

O

Q

PA B

O

(2) Let P be a point on the minor arc AB of a circle (O).

∠AOB + 2∠APB = 360◦.

(3) The opposite angles of a cyclic quadrilateral are supplementary.(4) The angle contained in a semicircle is a right angle.(5) Equal chords subtend equal angles at the center.

4.2 Tests of congruence of triangles xix

P

A BO

A

B D

O

C

4.2 Tests of congruence of triangles

4.2.1 Construction of a triangle with three given elements

A triangle has six elements: three sides and three angles. Consider the construc-tion of a triangle given three of its six elements. The triangle is unique (up to sizeand shape) if the given data are in one of the following patterns.

(1) SSSGiven three lengths a, b, c, we construct a segment BC with lengtha, and the two circles B(c) and C(b). These two circles intersect at two points if(and only if) b + c > a [triangle inequality]. There are two possible positions ofA. The resulting two triangles are congruent.

(2) SAS Given b, c, and angle A < 180◦, the existence and uniqueness oftriangle ABC is clear.

(3) ASA or AAS These two patterns are equivalent since knowing two of theangles of a triangle, we easily determine the third (their sum being 180◦). GivenB, C and a, there is clearly a unique triangle provided B + C < 180◦.

(4) RHS Given a, c and C = 90◦.

4.2.2 Congruence tests

These data patterns also provide the valid tests of congruence of triangles. Two tri-angles ABC and XY Z are congruent if their correspondingelements are equal.Two triangles are congruent if they have three pairs of equal elements in one thefive patterns above. The five valid tests of congruence of triangles are as follows.

xx Some basic principles

(1) SSS: �ABC ≡ �XY Z if

AB = XY, BC = Y Z, CA = ZX.

A

B C

X

Y

Z

(2) SAS: �ABC ≡ �XY Z if

AB = XY, ∠ABC = ∠XY Z, BC = Y Z.

A

B C

X

Y

Z

(3) ASA: �ABC ≡ �XY Z if

∠BAC = ∠Y XZ, AB = XY, ∠ABC = ∠XY Z.

A

B C

X

Y

Z

(4) AAS. We have noted that this is the same as ASA: �ABC ≡ �XY Z if

∠BAC = ∠Y XZ, ∠ABC = ∠XY Z, BC = Y Z, .

4.2 Tests of congruence of triangles xxi

A

B C

X

Y

Z

(5) RHS. The ASS is not a valid test of congruence. Here is an example. Thetwo triangles ABC andXY Z are not congruent even though

∠BAC = ∠Y XZ, AB = XY, BC = Y Z.

A X

B Y

C Z

However, if the equal angles are right angles, then the third pair of sides areequal:

AC2 = BC2 −AB2 = Y Z2 −XY 2 = XZ2,

and AC = XZ. The two triangles are congruent by the SSS test. Without repeat-ing these details, we shall simply refer to this as the RHS test. �ABC ≡ �XY Zif

∠BAC = ∠Y XZ = 90◦, BC = Y Z, AB = XY.

A X

B Y

C Z

The congruence tests for triangles form a paradigm for proofs in euclideangeometry. We shall illustrate this with numerous examples in Chapter 6. Here, wegive only one example, the converseof the Pythagorean theorem.

xxii Some basic principles

Theorem 4.1 (Converse of Pythagoras’ theorem).If the lengths of the sides of�ABC satisfy a2 + b2 = c2, then the triangle has a right angle at C.

ac

b

a

bC X

B Y

A Z

Proof. Consider a right triangle XY Z with ∠Z = 90◦, Y Z = a, and XZ = b.By the Pythagorean theorem, XY 2 = Y Z2 + XZ2 = a2 + b2 = c2 = AB2. Itfollows that XY = AB, and �ABC ≡ �XY Z by the test, and∠C = ∠Z = 90◦.

Chapter 5

Circumcircle and incircle

5.1 Circumcircle

The perpendicular bisector of a segment is the line perpendicular to it throughits midpoint.

5.1.1 The perpendicular bisector locus theorem

Theorem 5.1. A point P is equidistant from B and C if and only if P lies on theperpendicular bisector of BC.

P

B CM

Proof. LetM be the midpoint of BC.(⇒) If PB = PC, then �PBM ≡ �PCM by the test. This

means ∠PMB = ∠PMC and PM ⊥ BC. The point P is on the perpendicularbisector of BC.

(⇐) If P is on the perpendicular bisector of BC, then �PMB ≡ PMC bythe test. It follows that PB = PC.

xxiv Circumcircle and incircle

5.1.2 Construction of circumcircle

Given a triangle ABC, there is a circle (O) through the three vertices. The centerO of the circle, being equidistant from B and C, must lie on the perpendicularbisector of BC. For the same reason, it also lies on the perpendicular bisectors ofCA and AB. This is called the circumcenter of triangle ABC.

O

A

B C

5.1.3 Circumcircle of a right triangle

To construct a circle through the vertices of a right triangle ABC, mark the mid-pointM of the hypotenuse AB. The circleM(C) passes through A and B, and isthe circumcircle of the right triangle.

A B

C

M

5.2 The incircle xxv

5.2 The incircle

5.2.1 The angle bisector locus theorem

The distance from a point P to a line � is the distance between P and its pedal(orthogonal projection) on �.

P

K

A H

Theorem 5.2. A point P is equidistant from two lines if and only if P lies on thebisector of an angle between the two lines.

Proof. Let A be the intersection of the two lines, and H , K the pedals of a pointP on these lines.

(⇒) If PH = PK then �APH ≡ �APK by the test. Itfollows that ∠PAH = ∠PAK, and P lies on the bisector of angle HAK.

(⇐) If ∠PAH = ∠PAK, then �APH ≡ �APK by thetest. It follows that PH = PK.

5.2.2 Construction of incircle

The incircle of a triangle is the one which is tangent to each of the three sidesof the triangle. Its center is the common point of the angle bisectors, and can belocated by constructing two of them. This incenter I is equidistant from the threesides.

I

A

B C

xxvi Circumcircle and incircle

5.2.3 The incircle of a right triangle

If d is the diameter of the incircle of a right triangle, then

a+ b = c+ d.

C

B

Ab

a cd

5.3 Tangents of a circle

5.3.1 Tangent at a point on the circle

A tangent to a circle is a line which intersects the circle at only one point. Givena circle O(A), the tangent to a circle at A is the perpendicular to the radius OAat A.

A

O

A

O

B

PM

5.3.2 The tangents from a point to a circle

If P is a point outside a circle (O), there are two lines through P tangent to thecircle. Construct the circle with OP as diameter to intersect (O) at two points.These are the points of tangency.

The two tangents have equal lengths since �OAP ≡ �OBP by thetest.

5.4 Some interesting examples on the circumcircle and incircle of a trianglexxvii

5.4 Some interesting examples on the circumcircleand incircle of a triangle

(1) Let s = 12(a + b + c) be the semiperimeterof triangle ABC. The lengths of

the tangents from the vertices to the incircle are as follows.

tangent from A =s− a,tangent from B =s− b,tangent from C =s− c.

C A

B

b

a c

r

rr

(2) The following rearrangement shows that for an arbitrary triangle, the radiusof its incircle is given by

r =area of triangle

semi-perimeter of triangle.

r

xxviii Circumcircle and incircle

(3∗) Construct the circumcircle (O) of triangle ABC and select an arbitrarypoint P on it. Construct the reflectionsof P in the sidelines of the given triangle.These three reflection points always lie on a straight line �. Furthermore, as Pvaries on (O), the line � passes through a fixed point.

P

H

A

B C

(4∗) Construct the circumcircle (O) and incircle (I) of a triangle ABC. Selectan arbitrary point P on (I) and construct the perpendicular to IP at P . This is atangent to (I). Let it intersect the circle (O) at Y and Z.

Construct the tangents from Y and Z to (I). These two tangents always inter-sect on the circumcircle (O).

I

P

X

Y

ZO

A

B C

Chapter 6

Uses of congruence tests

6.1 Isosceles triangles

An isosceles triangle is one with two equal sides.

Proposition 6.1. A triangle is isosceles if and only if it has two equal angles.

B

A

CM B

A

CD

Proof. (⇒) Let ABC be a triangle in which AB = AC. Let M be the midpointof BC. Then �ABM ≡ �ACM by the test. From this concludethat ∠BAM = ∠CAM .

(⇐) Let ABC be a triangle in which ∠B = ∠C. Construct the perpendicularfrom A to BC, and let it intersect BC at D. Then �ABD ≡ �ACD by the

test. From this AB = AC and the triangle is isosceles.

xxx Uses of congruence tests

6.2 Chords of a circle

(1) LetM be a point on a chord BC of a circle (O). M is the midpoint of BC ifand only if OM ⊥ BC.

O

MB C

Proof. (⇒) IfM is the midpoint ofBC, then �OMB ≡ �OMC by thetest. Therefore, ∠OMB = ∠OMC and OM ⊥ BC.

(⇐) If OM ⊥ BC, then �OMB ≡ �OMC by the test. Itfollows that BM = CM , andM is the midpoint of BC.

(2) Equal chords of a circle are equidistant from the center. Conversely, chordsequidistant from the center are equal in length.

Proof. Suppose BC and PQ are equal chords of a circle (O), with midpointsM and N respectively. Then BM = PN , and �OBM ≡ �OPN by the

test. Therefore, OM = ON , and the chords are equidistant fromthe center O.

Conversely, if OM = ON , then �OBM ≡ �OPN by thetest. It follows that BM = PN , and BC = 2 · BM = 2 · PN = PQ.

6.3 Parallelograms xxxi

6.3 Parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides.

6.3.1 Properties of a parallelogram

(1) The opposite sides of a parallelogram are equal in length. The opposite anglesof a parallelogram are equal.

A B

CD

Proof. Consider a parallelogram ABCD. Construct the diagonal BD. Note that�ABD ≡ �CDB by the test. It follows that AB = CD andAD = CB, and ∠BAD = ∠DCB. Similarly, ∠ABC = ∠CDA.

(2) The diagonals of a parallelogram bisect each other.

M

A B

CD

Proof. Consider a parallelogram ABCD whose diagonals AC and BD intersectat M . Note that AD = CB and �AMD ≡ �CMB by the test.It follows thatAM = CM andDM = BM . The diagonals bisect each other.

xxxii Uses of congruence tests

6.3.2 Quadrilaterals which are parallelograms

(1) If a quadrilateral has two pairs of equal opposite sides, then it is a parallelo-gram.

A B

CD

Proof. Let ABCD be a quadrilateral in which AB = CD and AD = BC. Con-struct the diagonal BD. Then �ABD ≡ �CDB by the test. Itfollows that and AD//BC. Also, andAB//DC. Therefore, ABCD is a parallelogram.

(2) If a quadrilateral has one pair of equal and parallel sides, then it is a paral-lelogram.

A B

CD

Proof. Suppose ABCD is a quadrilateral in which AB//DC and AB = CD.Then �ABD ≡ �CDB by the test. It follows that AD = CBandABCD is a parallelogram since .

6.3.3 Special parallelograms

A quadrilateral is(1) a rhombus if its sides are equal,(2) a rectangle if its angles are equal (and each 90◦),(3) a square if its sides are equal and its angles are equal.

6.4 The midpoint theorem and its converse xxxiii

These are all parallelograms since(1) a rhombus has two pairs of sides, and(2) a rectangle has two pairs of sides.Clearly, then a square is a parallelogram.

(1) A parallelogram is a rhombus if and only if its diagonals are perpendicularto each other.

(2) A parallelogram is a rhombus if and only if it has an angle bisected by adiagonal.

(3) A parallelogram is a rectangle if and only if its diagonals are equal inlength.

6.4 The midpoint theorem and its converse

6.4.1 The midpoint theorem

Given triangle ABC, let E and F be the midpoints of AC and AB respectively.The segment FE is parallel to BC and its length is one half of the length of BC.

A

B C

FE

D

Proof. Extend FE to D such that FE = ED. Note that �CDE ≡ �AFE bythe test. It follows that(i) CD = AF = BF ,(ii) ∠CDE = ∠AFE, and CD//BA. Therefore, CDFB is a parallelogram, andFE//BC. Also, BC = FD = 2FE.

xxxiv Uses of congruence tests

6.4.2 The converse of the midpoint theorem

Let F be the midpoint of the side AB of triangle ABC. The parallel through Fto BC intersects AC at its midpoint.

A

B C

FE

D

Proof. Construct the parallel through C to AB, and extend FE to intersect thisparallel atD. Then, CDFB is a parallelogram, and CD = BF = FA. It followsthat �AEF ≡ CED by the test. This means that AE = CE, andE is the midpoint of AC.

6.4.3 Why are the three medians concurrent?

Let E and F be the midpoints of AC and AB respectively, and G the intersectionof the mediansBE and CF .

Construct the parallel through C to BE, and extend AG to intersect BC atD,and this parallel at H . .

A

B C

FE

D

G

H

6.5 Some interesting examples xxxv

By the converse of the midpoint theorem,G is the midpoint ofAH , andHC =2 ·GE

Join BH . By the midpoint theorem, BH//CF . It follows that BHCG is aparallelogram. Therefore, D is the midpoint of (the diagonal) BC, and AD isalso a median of triangle ABC. We have shown that the three medians of triangleABC intersect at G, which we call the centroid of the triangle.

Furthermore,

AG =GH = 2GD,

BG =HC = 2GE,

CG =HB = 2GF.

The centroid G divides each median in the ratio 2 : 1.

6.5 Some interesting examples

(1) Why are the three altitudes of a triangle concurrent?LetABC be a given triangle. Through each vertex of the triangle we construct

a line parallel to its opposite side. These three parallel lines bound a larger triangleA′B′C ′. Note that ABCB ′ and ACBC ′ are both parallelograms since each hastwo pairs of parallel sides. It follows that B ′A = BC = AC ′ and A is themidpoint of B ′C ′.

H

X

Y

Z

A

B C

A′

B′C′

xxxvi Uses of congruence tests

Consider the altitude AX of triangle ABC. Seen in triangle A′B′C ′, this lineis the perpendicular bisector of B ′C ′ since it is perpendicular to B ′C ′ through itsmidpoint A. Similarly, the altitudes BY and CZ of triangle ABC are perpendic-ular bisectors of C ′A′ and A′B′. As such, the three lines AX , BY , CZ concur ata pointH . This is called the orthocenter of triangle ABC.

(2∗) The butterfly theorem. Let M be the midpoint of a chord PQ of a circle(O). AB and CD are two chords passing throughM . Join BC to intersect PQ atX and AD to intersect PQ at Y . ThenMX =MY .

MP Q

O

A

B

C

D

Y X

Chapter 7

The regular hexagon, octagon anddodecagon

A regular polygon of n sides can be constructed (with ruler and compass) if it ispossible to divides the circle into n equal parts by dividing the 360◦ at the centerinto the same number of equal parts. This can be easily done for n = 6 (hexagon)and 8 (octagon). 1 We give some easy alternatives for the regular hexagon, octagonand dodecagon.

7.1 The regular hexagon

A regular hexagon can be easily constructed by successively cutting out chords oflength equal to the radius of a given circle.

1Note that if a regular n-gon can be constructed by ruler and compass, then a regular 2n-goncan also be easily constructed.

xxxviii The regular hexagon, octagon and dodecagon

7.2 The regular octagon

(1) Successive completion of rhombi beginning with three adjacent 45◦-rhombi.

(2) We construct a regular octagon by cutting from each corner of a givensquare (of side length 2) an isosceles right triangle of (shorter) side x. This means2 − 2x : x =

√2 : 1, and 2 − 2x =

√2x; (2 +

√2)x = 2,

x =2

2 +√

2=

2(2 −√2)

(2 +√

2)(2 −√2)

= 2 −√

2.

P

Q

A B

CD

x 2 − 2x x

x

O

P

Q

A B

CD

Therefore AP = 2−x =√

2, which is half of the diagonal of the square. Thepoint P , and the other vertices, can be easily constructed by intersecting the sidesof the square with quadrants of circles with centers at the vertices of the squareand passing through the center O.

7.3 The regular dodecagon xxxix

7.3 The regular dodecagon

(1) Trisection of right angles.

(2) A regular dodecagoncan be formed from 4 equilateral triangles inside asquare. 8 of its vertices are the intersections of the sides of these equilateral trian-gles, while the remaining 4 are the midpoints of the sides of the squares formedby the vertices of the equilateral triangles inside the square.

An easy dissection shows that the area of the regular dodecagon is 34

of that ofthe (smaller) square containing it.

xl The regular hexagon, octagon and dodecagon

(3) Successive completion of rhombi beginning with five adjacent 30◦-rhombi.

This construction can be extended to other regular polygons. If an angle ofmeasure θ := 360

2n

◦can be constructed with ruler and compass, beginning with

n − 1 adjacent θ-rhombi, by succesively completing rhombi, we obtain a regular2n-gons tesellated by

(n− 1) + (n− 2) + · · ·+ 2 + 1 =1

2n(n− 1)

rhombi.

Chapter 8

Similar triangles

8.1 Tests for similar triangles

Two triangles are similar if their corresponding angles are equal and their corre-sponding sides proportional to each other, i.e., �ABC ∼ �XY Z if and onlyif

∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z,and

BC

Y Z=CA

ZX=AB

XY.

It is enough to conclude similarity of two triangles when they have three pairsof

(1) AAA : if the triangles have two (and hence three) pairs of equal angles.

A

B C

X

YZ

xlii Similar triangles

(2) proportional SAS: ifCA

ZX=AB

XYand ∠A = ∠X .

A

B C

X

YZ

(3) proportional SSS: ifBC

Y Z=CA

ZX=AB

XY.

A

B C

X

YZ

8.2 The parallel intercepts theorem xliii

8.2 The parallel intercepts theorem

The intercepts on two transversals between a set of parallel lines are proportionalto each other: if �0, �1, �2, . . . , �n−1, �n are parallel lines and L1 and L2 are twotransversals intersecting these lines at A0, A1, A2, . . . , An−1, An, and B0, B1, B2,. . . , Bn−1, Bn, then

A0A1

B0B1

=A1A2

B1B2

= · · · =An−1AnBn−1Bn

.

A0 B0

A1 B1

A2 B2

An−1 Bn−1

An Bn

C0

C1

Cn−1

L1L2

�0

�1

�2

�n−1

�n

Proof. Through B1, B2, . . . , Bn construct lines parallel to L1, intersecting �0, �1,. . . , �n−1 at C0, C1, . . . , Cn−1 respectively. Note that(i) B1C0 = A1A0, B2C1 = A2A1, . . . , BnCn−1 = AnAn−1, and(ii) the triangles B0B1C0, B1B2C1, . . . , Bn−1BnCn−1 are similar.It follows that

B1C0

B0B1

=B2C1

B1B2

= · · · =BnCn1

Bn−1Bn.

Hence,A0A1

B0B1=A1A2

B1B2= · · · =

An−1AnBn−1Bn

.

xliv Similar triangles

8.3 The angle bisector theorem

The bisector of an angle divides the opposite side in the ratio of the remaining twosides, i.e.,if AX bisects angle A, then

BX

XC=AB

AC.

A

B CX

D

Proof. Construct the parallel through B to the bisector to intersect the extensionof CA atD. Note that

∠ABD = ∠BAX = ∠CAX = ∠CDB = ∠ADB.

This means that triangle ABD is isosceles, and AB = AD. Now, from theparallelism of AX and DB, we have

AX

XB=DA

AC=AB

AC.

8.4 Some interesting examples of similar triangles xlv

8.4 Some interesting examples of similar triangles

(1) The altitude from the right angle vertex divides a right triangle into two trian-gles each similar to the given right triangle.

A B

C

D

Proof. Let ABC be a triangle with a right angle at C. IfD is a point on AB suchthatDC ⊥ AB, then �ACD ∼ �CBD ∼ �ABC.

Here are some interesting consequences.

(a) From �ACD ∼ �CBD, we haveAD

CD=CD

BD. It follows that

CD2 = AD · CD.

(b) From �CBD ∼ �ABC, we haveBD

CB=BC

ABso that

BC2 = BD · AB.

Similarly, from �ACD ∼ �ABC, we haveAD

AC=AC

ABso that

AC2 = AD · AB.

It follows that

BC2 +AC2 = BD ·AB +AD ·AB = (BD +AD) ·AB = AB ·AB = AB2.

This gives an alternative proof of the Pythagorean theorem.(a) and (b) above give simple constructions of geometric means.

xlvi Similar triangles

Construction 8.1. Given a point P on a segment AB, construct(1) a semicircle with diameter AB,(2) the perpendicular to AB at P , intersecting the semicircle at Q.

ThenPQ2 = AP × PB.

A BO P

Q

A BO P

Q

θ

θ

(2) The radius of the semicircle is the geometric mean of AP amd BQ.

A

B

O

Q

P

This follows from the similarity of the right triangles OAP and QBO. If

r = OA = OB, thenr

QB=AP

r. From this, r2 = AP · BQ.

8.4 Some interesting examples of similar triangles xlvii

(3∗) Heron’s formula for the area of a triangle.

I

I′

Z′

s− b s− c s− a

r′

r

Y ′ Y

B

C A

The A-excircle of triangle ABC is the circle which is tangent to the side BCand to the extensions of AB and AC. If Y ′ and Z ′ are the points of tangency withAC and AB, it is easy to see that AY ′ = AZ ′ = s, the semiperimeter. Hence,CY ′ = s− b. If the incircle touches AC at Y , we have known from Example 1 of§5.4 that CY = s− c and AY = s− a.

From these we can find the radii r of the incircle, and r′ of theA-excircle quiteeasily. From the similarity of triangles AIY and AI ′Y ′, we have

r

r′=s− as.

Note that also �ICY ∼ �CI ′Y ′, from which we haver

s− b =s− cr′. and

r · r′ = (s− b)(s− c).Multiplying these two equations together, we obtain

r2 =(s− a)(s− b)(s− c)

s.

From Example 2 of §5.4, the area of triangle ABC is given by � = rs. Hence,we have the famous Heron formula

� =√s(s− a)(s− b)(s− c).

xlviii Similar triangles

Chapter 9

Tangency of circles

9.1 External and internal tangency of two circles

Two circles (O) and (O′) are tangent to each other if they are tangent to a line� at the same line P , which is a common point of the circles. The tangency isinternal or external according as the circles are on the same or different sides ofthe common tangent �.

O O′ O O′P

P

� �

The line joining their centers passes through the point of tangency.The distance between their centers is the sum or difference of their radii, ac-

cording as the tangency is external or internal.

l Tangency of circles

9.2 Three mutually tangent circles

Given three non-collinear points A, B, C, construct(i) triangle ABC and its incircle, tangent to BC, CA, AB respectively at X , Y ,Z,(ii) the circles A(Z), B(Y ) and C(X).These are tangent to each other externally.

C A

B

Z

Y

X

9.3 Some interesting examples of tangent circles

(1) In each of the following diagrams, the shaded triangle is similar to the 3−4−5triangle.

9.3 Some interesting examples of tangent circles li

lii Tangency of circles

(2∗) Given triangle ABC, letD, E, F be the midpoints of the sides BC, CA,AB respectively. The circumcircle of triangle DEF is always tangent internallyto the incircle of triangle ABC.

I

D

EF

A

B C

Chapter 10

The regular pentagon

10.1 The golden ratio

The construction of the regular pentagon is based on the division of a segmentin the golden ratio. 1 Given a segment AB, to divide it in the golden ratio is toconstruct a point P on it so that the area of the square on AP is the same as thatof the rectangle with sides PB and AB, i.e.,

AP 2 = AB · PB.

A construction of P is shown in the second diagram.

A BP BA

M

P

1In Euclid’s Elements, this is called division into the extreme and mean ratio.

liv The regular pentagon

Suppose PB has unit length. The length ϕ of AP satisfies

ϕ2 = ϕ+ 1.

This equation can be rearranged as

(ϕ− 1

2

)2

=5

4.

Since ϕ > 1, we have

ϕ =1

2

(√5 + 1

).

Note thatAP

AB=

ϕ

ϕ+ 1=

1

ϕ=

2√5 + 1

=

√5 − 1

2.

This explains the construction above.

10.2 The diagonal-side ratio of a regular pentagon

Consider a regular pentagon ACBDE. It is clear that the five diagonals all haveequal lengths. Note that(1) ∠ACB = 108◦,(2) triangle CAB is isosceles, and(3) ∠CAB = ∠CBA = (180◦ − 108◦) ÷ 2 = 36◦.

In fact, each diagonal makes a 36◦ angle with one side, and a 72◦ angle withanother.

A

E D

BP

C

10.3 Construction of a regular pentagon with a given diagonal lv

It follows that(4) triangle PBC is isosceles with ∠PBC = ∠PCB = 36◦,(5) ∠BPC = 180◦ − 2 × 36◦ = 108◦, and(6) triangles CAB and PBC are similar.

Note that triangle ACP is also isosceles since(7) ∠ACP = ∠APC = 72◦. This means that AP = AC.

Now, from the similarity of CAB and PBC, we have AB : AC = BC : PB.In other words AB · AP = AP · PB, or AP 2 = AB · PB. This means that Pdivides AB in the golden ratio.

10.3 Construction of a regular pentagon with a givendiagonal

Given a segment AB, we construct a regular pentagon ACBDE with AB as adiagonal.(1) Divide AB in the golden ratio at P .(2) Construct the circles A(P ) and P (B), and let C be an intersection of thesetwo circles.(3) Construct the circles A(AB) and B(C) to intersect at a point D on the sameside of BC as A.(4) Construct the circles A(P ) and D(P ) to intersect at E.

Then ACBDE is a regular pentagon with AB as a diameter.

A

E D

BP

C

lvi The regular pentagon

10.4 Various constructions of the regular pentagon

(1) To construct a regular pentagon with vertices on a given circle O(A),(i) construct a radius OP perpendicular to OA,(ii) mark the midpointM of OP and join it to A,(iii) bisect the angle OMA and let it intersect OA at Q,(iv) construct the perpendicular to OA at Q to intersect the circle at B and E,(v) mark C, D on the circle such that AB = BC and AE = ED.

Then ABCDE is a regular pentagon.

E

D C

B

A

PM

Q

O

(2) A compass-only costruction of the regular pentagon:2 construct(i) (O) = O(A),(ii) A(O) to mark B and F on (O),(iii) B(O) to mark C,(iv) C(O) to mark D,(v) D(O) to mark E,(vi) A(C) and D(AC) to intersect atX ,(vii) Y = C(OX) ∩E(OX) inside (O),(viii) P = A(OY ) ∩ (O),(ix) Q = P (A) ∩ (O),(x) D(AP ) to intersect (O) at R and S (so that R is between Q and D),(xi) S(R) to intersect (O) at T .

Then AQRST is a regular pentagon inscribed in (O).

2Modification of a construction given by E. P. Starke, 3 of a regular pentagon in 13 steps usingcompass only.

10.4 Various constructions of the regular pentagon lvii

A

BC

D

E F

O

X

Y

P

Q

R

S

T

(3∗) Another construction of an inscribed regular pentagon. Let O(A) be agiven circle.(i) Construct an isosceles triangle AXY whose height is 5

4of the radius of the

circle.(ii) Construct A(O) intersecting O(A) at B ′ and C ′.(iii) Mark P = AX ∩OB′ and Q = AY ∩OC ′.(iv) Join P and Q by a line intersecting the given circle at B and C.

Then AB and AC are two sides of a regular pentagon inscribed in the circle.

C′B′

YX

QPCB

O

A

lviii The regular pentagon

10.5 Some interesting constructions of the golden ra-tio

(1) Construction of 36◦, 54◦, and 72◦ angles. Each of the following constructionsbegins with the division of a segment AB in the golden ratio at P .

36◦A

B

C

P36◦ 36◦

36◦A

B

C

D

P

54◦

54◦

AB

C

P72◦

72◦

(2∗) Odom’s construction. LetD and E be the midpoints of the sides AB andAC of an equilateral triangle ABC. If the line DE intersects the circumcircle ofABC at F , then E dividesDF in the golden ratio.

A

B

D

C

EF

10.5 Some interesting constructions of the golden ratio lix

(3) Given a segment AB, erect a square on it, and an adjacent one with baseBC. If D is the vertex above A, construct the bisector of angle ADC to intersectAB at P . Then P divides AB in the golden ratio.

A B C

D

P

Proof. If AB = 1, then

AP : PC =1 :√

5,

AP : AC =1 :√

5 + 1

AC : 2AB =1 :√

5 + 1

AB : AP =√

5 + 1 : 2

=ϕ : 1.

This means that P divides AB in the golden ratio.

(4) Hofstetter’s parsimonious construction. 4 Given a segment AB, construct(i) C1 = A(B),(ii) C2 = B(A), intersecting C1 at C and D,(iii) the line AB to intersect C1 at E (apart from B),(iv) C3 = E(B) to intersect C2 at F (so that C and F are on opposite sides ofAB),(v) the segment CF to intersect AB at G.

Then the pointG divides the segment AB in the golden section.

4K. Hofstetter, Another 5-step division of a segment in the golden section, Forum Geom., 4(2004) 21–22.

lx The regular pentagon

A B

C

D

E

F

G

C1

C2

C3

(5) Hofstetter’s rusty compass construction. 5

(i) Construct A(B) and B(A) intersecting at C and D.(ii) Join CD to intersect AB at its midpointM .(iii) ConstructM(AB) to intersect B(A) at E on the same side of AB as C.(iv) JoinDE to intersect AB at P .

Then P divides the segment in the golden ratio.

A B

C

D

M P

E

5K. Hofstetter, Division of a segment into golden ratio with ruler and rusty compass, ForumGeom., 5 (2005) 133–134.