GeometryClass Examples (July 29)

Paul Yiu

Department of MathematicsFlorida Atlantic University

a

bc

A

B C

Summer 2014

1

The Pythagorean Theorem

Theorem (Pythagoras). The lengths a ≤ b < c of the sides of a right trianglesatisfy the relation

a2 + b2 = c2.

Proof.

a b

a

b

ab

a

b

4 1

23

4

1

2

3a

a

b

b

Theorem (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.It follows that XY = AB, and �ABC ≡ �XY Z by the SSS test, and∠C = ∠Z = 90◦.

2

Theorem (The law of sines). Let R denote the circumradius of a triangleABC.

2R =a

sinα=

b

sin β=

c

sin γ.

O

C

A

BD

EFO

C

A

BD

α

α

Since the area of a triangle is given by Δ = 12bc sinα, the circumradius

can be written as

R =abc

4Δ.

3

The law of cosines

Given a triangle ABC, we denote by a, b, c the lengths of the sides BC,CA, AB respectively.

Theorem (The law of cosines).

c2 = a2 + b2 − 2ab cos γ.

a

bc

A

B CX a

b

c

A

B C X

Proof. Let AX be the altitude on BC.

c2 = BX2 + AX2

= (a− b cos γ)2 + (b sin γ)2

= a2 − 2ab cos γ + b2(cos2 γ + sin2 γ)

= a2 + b2 − 2ab cos γ.

4

Apollonius’ Theorem

Theorem. Given triangle ABC, let D be the midpoint of BC. The length ofthe median AD is given by

AB2 + AC2 = 2(AD2 +BD2).

A

B CD

Proof. Applying the law of cosines to triangles ABD and ACD, and notingthat cosADB = − cosADC, we have

AB2 = AD2 +BD2 − 2AD · BD · cosADB;

AC2 = AD2 + CD2 − 2AD · CD · cosADC,

AC2 = AD2 +BD2 + 2AD · BD · cosADB.

The result follows by adding the first and the third lines.

If ma denotes the length of the median on the side BC,

m2a =

1

4(2b2 + 2c2 − a2).

5

4.1. Angle bisector theorem

Theorem (Angle bisector theorem). The bisectors of an angle of a triangledivide its opposite side in the ratio of the remaining sides. If AX and AX ′

respectively the internal and external bisectors of angle BAC, then BX :XC = c : b and BX ′ : X ′C = c : −b.

c b

Z′

Z

A

B CX X′

Proof. Construct lines through C parallel to the bisectors AX and AX ′ tointersect the line AB at Z and Z ′.

(1) Note that ∠AZC = ∠BAX = ∠XAC = ∠ACZ. This meansAZ = AC. Clearly, BX : XC = BA : AZ = BA : AC = c : b.

(2) Similarly, AZ ′ = AC, and BX ′ : X ′C = BA : AZ ′ = BA :−AC = c : −b.

6

The lengths of the bisectors

Theorem. (a) The lengths of the internal and external bisectors of angle A

are respectively

ta =2bc

b+ ccos

α

2and t′a =

2bc

|b− c| sinα

2.

c bta t′a

A

B CX X′

Remarks. (1) 2bcb+c is the harmonic mean of b and c. It can be constructed as

follows. If the perpendicular to AX at X intersects AC and AB at Y andZ, then AY = AZ = 2bc

b+c .

c b

ta

Y

Z

A

BC

X

(2) Applying Stewart’s Theorem with λ = c and μ = ±b, we also obtainthe following expressions for the lengths of the angle bisectors:

t2a = bc

(1−

(a

b+ c

)2),

t′2a = bc

((a

b− c

)2

− 1

).

7

The incircle

The internal angle bisectors of a triangle are concurrent at the incenter ofthe triangle. This is the center of the incircle, the circle tangent to the threesides of the triangle.

Let the bisectors of angles B and C intersect at I . Consider the pedals ofI on the three sides. Since I is on the bisector of angle B, IX = IZ. SinceI is also on the bisector of angle C, IX = IY . It follows IX = IY = IZ,and the circle, center I , constructed through X , also passes through Y andZ, and is tangent to the three sides of the triangle.

I

X

Y

Z

A

B C s − b s − c

s − c

s − a

s − a

s − b

I

X

Y

Z

A

B C

This is called the incircle of triangle ABC, and I the incenter.Let s be the semiperimeter of triangle ABC. The incircle of triangle

ABC touches its sides BC, CA, AB at X , Y , Z such that

AY =AZ = s− a,

BZ =BX = s− b,

CX =CY = s− c.

The inradius of triangle ABC is the radius of its incircle. It is given by

r =2Δ

a+ b+ c=

Δ

s.

8

The excircles

The internal bisector of each angle and the external bisectors of the remain-ing two angles are concurrent at an excenter of the triangle. An excircle canbe constructed with this as center, tangent to the lines containing the threesides of the triangle.

Z

X

Y

ra

ra

ra

Ic

Ib

Ia

C

A

B

The exradii of a triangle with sides a, b, c are given by

ra =Δ

s− a, rb =

Δ

s− b, rc =

Δ

s− c.

The areas of the triangles IaBC, IaCA, and IaAB are 12ara, 1

2bra, and 12cra

respectively. Since

Δ = −ΔIaBC +ΔIaCA+ΔIaAB,

we have

Δ =1

2ra(−a+ b+ c) = ra(s− a),

from which ra =Δs−a .

9

Heron’s formula for the area of a triangle

Consider a triangle ABC with area Δ. Denote by r the inradius, and ra theradius of the excircle on the side BC of triangle ABC. It is convenient tointroduce the semiperimeter s = 1

2(a+ b+ c).

Ia

YY ′

I

A

B

C

ra r

(1) From the similarity of triangles AIY and AI ′Y ′,

r

ra=

s− a

s.

(2) From the similarity of triangles CIY and I ′CY ′,

r · ra = (s− b)(s− c).

(3) From these,

r =

√(s− a)(s− b)(s− c)

s,

ra =

√s(s− b)(s− c)

s− a.

Theorem (Heron’s formula).

Δ =√

s(s− a)(s− b)(s− c).

Proof. Δ = rs.

10

Menelaus’ theorem

Theorem (Menelaus). Given a triangle ABC with points X , Y , Z on theside lines BC, CA, AB respectively, the points X , Y , Z are collinear if andonly if

BX

XC· CY

Y A· AZZB

= −1.

A

B CX

Y

Z

W

Proof. (=⇒) Let W be the point on AC such that BW//XY . Then,

BX

XC=

WY

Y C, and

AZ

ZB=

AY

YW.

It follows that

BX

XC· CY

Y A· AZZB

=WY

Y C· CY

Y A· AYYW

=CY

Y C· AYY A

· WY

YW= −1.

(⇐=) Suppose the line joining X and Z intersects AC at Y ′. Fromabove,

BX

XC· CY ′

Y ′A· AZZB

= −1 =BX

XC· CY

Y A· AZZB

.

It follows thatCY ′

Y ′A=

CY

Y A.

The points Y ′ and Y divide the segment CA in the same ratio. These mustbe the same point, and X , Y , Z are collinear.

11

Ceva’s theorem

Theorem (Ceva). Given a triangle ABC with points X , Y , Z on the sidelines BC, CA, AB respectively, the lines AX , BY , CZ are concurrent ifand only if

BX

XC· CY

Y A· AZZB

= +1.

P

X

Y

Z

A

B C

Proof. (=⇒) Suppose the lines AX , BY , CZ intersect at a point P . Con-sider the line BPY cutting the sides of triangle CAX . By Menelaus’ theo-rem,

CY

Y A· APPX

· XB

BC= −1, or

CY

Y A· PA

XP· BX

BC= +1.

Also, consider the line CPZ cutting the sides of triangle ABX . ByMenelaus’ theorem again,

AZ

ZB· BC

CX· XP

PA= −1, or

AZ

ZB· BC

XC· XP

PA= +1.

Multiplying the two equations together, we have

CY

Y A· AZZB

· BX

XC= +1.

(⇐=) Exercise.

1

Barycentric coordinates on a line

w vB C

X

Let B and C be two fixed points.A point X on the line BC is defined by one of the following:(1) the division ratio BX

XC = wv ,

(2) distributing a unit mass at B and C so as to balance at X:the absolute barycentric coordinates

X =vB + wC

v + w,

(3) the homogeneous barycentric coordinates relative to BC:X = v : w.

2

Barycentric coordinates with respect to a triangle

Given a reference triangle ABC, we put atthe vertices A, B, Cmasses u, v, w respectively,and determine the balance point.

A(u)

B(v) C(w)

w v

A(u)

B CX(v + w)

v + w

u

P (u+ v + w)

A

B CX

The masses at B and C can be replaced by a single mass v + w at the point

X =v · B + w · C

v + w.

Together with the mass at A,this can be replaced by a mass u+ v + wat the point P which divides AX in the ratioAP : PX = v + w : u.This is the point with absolute barycentric coordinates

u · A+ v · B + w · Cu+ v + w

,

provided u+ v + w �= 0.

We also say that the balance point P hashomogeneous barycentric coordinates (u : v : w)with reference to ABC.

13

Cevian triangle

Theorem (Ceva).Let X , Y , Z be points on the lines BC, CA, AB respectively.The lines AX , BY , CZ are collinearif and only ifthe given points have coordinates of the form

X = (0 : y : z),Y = (x : 0 : z),Z = (x : y : 0),

for some x, y, z. If this condition is satisfied, the common point of the linesAX , BY , CZ is P = (x : y : z).

A

B C

Z

X

Y

P

A

B C

CP

AP

BP

P

(1) The points X , Y , Z are called the traces of P .We also say that XY Z is the cevian triangle of P (with reference to trian-gle ABC).Sometimes, we shall adopt the more functional notation for the cevian tri-angle and its vertices:

cev(P ) : AP = (0 : y : z), BP = (x : 0 : z), CP = (x : y : 0).

(2) The point P divides the segment AX in the ratio

PX : AX = x : x+ y + z.

8

Theorem. The homogeneous barycentric coordinates of P are in the propor-tion of the (signed) areas of triangles PBC, PCA, PAB.

A

B C

CP

AP

BP

P P

A

B CX

Proof. Since PX : AX = x : x+ y + z,ΔPBX : ΔABX = x : x+ y + z, andΔPXC : ΔAXC = x : x+ y + z.Therefore,

ΔPBC = ΔPBX +ΔPXC

=x

x+ y + z·ΔABX +

x

x+ y + z·ΔAXC

=x

x+ y + z(ΔABX +ΔAXC)

=x

x+ y + z·ΔABC.

Similarly,

ΔPCA =y

x+ y + z·ΔABC, ΔPAB =

z

x+ y + z·ΔABC.

From these,ΔPBC : ΔPCA : ΔPAB = x : y : z.

10

Area and barycentric coordinates

Theorem. If for i = 1, 2, 3, Pi = xi ·A+yi ·B+zi ·C (in absolute barycentriccoordinates) , then the area of the oriented triangle P1P2P3 is

ΔP1P2P3 =

∣∣∣∣∣∣x1 y1 z1x2 y2 z2x3 y3 z3

∣∣∣∣∣∣ ·ΔABC.

9

Equations of straight lines

Two-point form

The area formula has an easy and extremely important consequence: theequation of the line joining two points with coordinates (x1 : y1 : z1) and(x2 : y2 : z2) is ∣∣∣∣∣∣

x1 y1 z1x2 y2 z2x y z

∣∣∣∣∣∣ = 0,

or(y1z2 − y2z1)x+ (z1x2 − z2x1)y + (x1y2 − x2y1)z = 0.

Examples

(1) The equations of the sidelines BC, CA, AB are respectively x = 0,y = 0, z = 0.

(2) Given a point P = (u : v : w), the cevian line AP has equationwy − vz = 0; similarly for the other two cevian lines BP and CP . Theselines intersect corresponding sidelines at the traces of P :

AP = (0 : v : w), BP = (u : 0 : w), CP = (u : v : 0).

(3) The equation of the line joining the centroid and the incenter is∣∣∣∣∣∣1 1 1a b cx y z

∣∣∣∣∣∣ = 0,

or (b− c)x+ (c− a)y + (a− b)z = 0.

10

Intersection of two lines

The intersection of the two lines

p1x+ q1y + r1z = 0,p2x+ q2y + r2z = 0

is the point(q1r2 − q2r1 : r1p2 − r2p1 : p1q2 − p2q1).

Proposition 0.1. Three lines pix+ qiy+ riz = 0, i = 1, 2, 3, are concurrentif and only if ∣∣∣∣∣∣

p1 q1 r1p2 q2 r2p3 q3 r3

∣∣∣∣∣∣ = 0.

11

The infinite point of a line

The infinite point of a line L has homogeneous coordinates given bythe difference of the absolute barycentric coordinates of two distinct pointson the line.As such, the coordinate sum of an infinite point is zero.We think of all infinite points constituting the line at infinity,L∞, which has equation x+ y + z = 0.

Examples

1. The infinite points of the side lines BC, CA, AB are (0 : −1 : 1),(1 : 0 : −1), (−1 : 1 : 0) respectively.

2. The infinite point of the A−altitude has homogeneous coordinates

(0 : SC : SB)− a2(1 : 0 : 0) = (−a2 : SC : SB).

3. More generally, the infinite point of the line px+ qy + rz = 0 is

(q − r : r − p : p− q).

12

Parallel lines

Parallel lines have the same infinite point.The line through P = (u : v : w) parallel to L : px + qy + rz = 0 hasequation ∣∣∣∣∣∣

q − r r − p p− qu v w

x y z

∣∣∣∣∣∣ = 0.

1. Find the equations of the lines through P = (u : v : w) parallel to theside lines.

2. Let DEF be the medial triangle of ABC, and P a point with ceviantriangle XY Z (with respect to ABC. Find P such that the lines DX ,EY , FZ are parallel to the internal bisectors of angles A, B, C respec-tively. 1

1The Nagel point P = (b+ c− a : c+ a− b : a+ b− c).

15

Solution to Example (5, 8, 9) triangle.

ABC is a triangle with a = 5, b = 8, and c = 9. Show that the Gergonnepoint lies on the line joining the midpoints of the sides BC and AC.

9

85

I

C

BA

Z

Y

X

D E

Ge

Solution. s = 11.CY = CZ = s− c = 2,AY = s− a = 6, BZ = s− b = 3.

Absolute barycentric coordinates: D = B+C2 , E = A+C

2 .

Homogeneous barycentric coordinates:

X = 0 : 2 : 3Y = 1 : 0 : 3

Ge = 1 : 2 : 3

Therefore Ge =A+2B+3C

6 .The Gergonne point lies on the line DE if and only ifGe = t ·D + (1− t)E for some t:

A+ 2B + 3C

6= t · B + C

2+ (1− t) · A+ C

2=

1− t

2A+

t

2B +

1

2C.

This is the case with t = 23 .

16

Example (7, 8, 9) triangle.

ABC is a triangle with a = 7, b = 8, and c = 9. Show that the Nagelpoint lies on the line joining the points of tangency of incircle with the sidesAB and AC.

7

89

A

B C

ZY

Z′

Na

Y ′

17

Solution to Example (7, 8, 9) triangle.

ABC is a triangle with a = 7, b = 8, and c = 9. Show that the Nagelpoint lies on the line joining the points of tangency of incircle with the sidesAB and AC.

7

89

A

B C

ZY

Z′

Na

Y ′

Solution. s = 12.BZ = s− b = 4, CY = s− c = 3;AZ = 4, AY = 3.

Absolute barycentric coordinates: Y = 3A+5C8 , Z = 4A+5B

9 .

Homogeneous barycentric coordinates:

Y ′ = 5 : 0 : 3Z ′ = 5 : 4 : 0

Na = 5 : 4 : 3

Therefore Na =5A+4B+3C

12 .The Nagel point lies on the line Y Z if and only ifNa = t · Y + (1− t)Z for some t:

5A+ 4B + 3C

12= t·3A+ 5C

8+(1−t)·4A+ 5B

9=

32− 5t

72A+

5(1− t)

9B+

5t

8C.

This is the case with t = 25 .