Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

Solutions to Exercises

1. Let X be on AB and Y be on AC such that AX/AB = AY/AC. Show that XY ‖ BC.

Solution. Let Y ′ be the point on line AC such that XY ′ ‖ BC.

A

B C

XY ′

Then ∠AXY ′ = ∠ABC and ∠AY ′X = ∠ACB, so triangles AXY ′ and ABC are similar. Hence,AX/AB = AY ′/AC. But we are given that AX/AB = AY/AC, so AY ′/AC = AY/AC. This meanspoints Y and Y ′ coincide, so XY ‖ BC.

2. In the diagram below, PQ = PR, ZX ‖ QY , QY ⊥ PR, and PQ is extended to W such thatWZ ⊥ PW .

P

Q

R

W

XY

Z

(a) Show that 4QWZ ∼ 4RXZ.

(b) Show that Y Q = ZX − ZW .

Solution. (a) Since PR = PQ, we have ∠R = ∠PQR = ∠WQZ. Since ZX ‖ QY , we have∠ZXR = ∠QY R = 90◦. So, ∠ZXR = ∠ZWQ, and we have 4QWZ ∼ 4RXZ by AA Similarity.

(b) First, we note that RQ = RZ −QZ, which looks a lot like the expression we want to prove. SinceXZ ‖ Y Q, we have ∠RY Q = ∠RXZ and ∠RQY = ∠RZX, so 4RY Q ∼ 4RXZ. This similaritygives us RZ/ZX = RQ/Y Q, so RQ = (RZ/ZX)(Y Q).

From 4QWZ ∼ 4RXZ in the last part, we have RZ/ZX = QZ/ZW , so QZ = (RZ/ZX)(ZW ).Substituting these into RQ = RZ −QZ gives

(RZ)(Y Q)

ZX= RZ − (RZ)(ZW )

ZX.

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

Multiplying this equation by ZX/RZ gives the desired Y Q = ZX − ZW .

3. The Angle Bisector Theorem. If D is on BC such that AD bisects ∠BAC, then AB/BD =AC/CD. Take a look here for hints.

Solution. Let E be the point on line AC so that BE is parallel to AD. Then triangle ∠AEB = ∠CADand ∠ABE = ∠BAD. But AD is the angle bisector of ∠BAC, so ∠CAD = ∠BAD, which means∠AEB = ∠ABE. Hence, triangle ABE is isosceles, with AB = AE.

A

B CD

E

Furthermore, since BE is parallel to AD, triangles EBC and ADC are similar, which means

CE

AC=BC

CD.

We can re-write this equation asAE +AC

AC=BD + CD

CD,

which simplifies toAE

AC+ 1 =

BD

CD+ 1.

Then AE/AC = BD/CD. But AE = AB, so AB/AC = BD/CD, which implies AB/BD = AC/CD.

4. Prove that SAS Similarity works without using the law of sines.

Solution. To prove SAS similarity, we must prove that if we have two triangles ABC and DEF suchthat AB/DE = AC/DF and ∠BAC = ∠EDF , then triangles ABC and DEF are similar.

Since ∠BAC = ∠EDF , we can shift triangle ABC to triangle A′B′C ′ in the plane, so that points A′

and D coincide, B′ lies on DE, and C ′ lies on DF .

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

B′ C ′

A′ = D

E F

Since DB′ = AB and DC ′ = AC, we have that DB′/DE = DC ′/DF . Hence, by Exercise 1, B′C ′ isparallel to EF . Then ∠DB′C ′ = ∠DEF and ∠DC ′B′ = ∠DFE, which means triangles DB′C ′ andDEF are similar, so triangles ABC and DEF are similar.

5. PA and BQ bisect angles ∠RPQ and ∠RQP , respectively. Given that RX ⊥ PA and RY ⊥ BQ,prove that XY ‖ PQ.

AB

P Q

R

XY

Solution. We extend−−→RY and

−−→RX to meet PQ at C and D, respectively, as shown.

AB

C DP Q

R

XY

Since ∠RPA = ∠APQ, PX = PX, and ∠RXP = ∠DXP , we have 4RXP ∼= 4DXP by ASACongruence. Similarly, we have 4CY Q ∼= 4RY Q. Therefore, RX = XD and RY = Y C, so X andY are midpoints of RD and RC, respectively.

So, we have RY/RC = RX/RD = 1/2, which gives us4RYX ∼ 4RCD by SAS Similarity. Therefore,we have ∠RYX = ∠RCD, so XY ‖ DC. Since DC is on the same line as PQ, we have XY ‖ PQ.

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

6. Chords AB and CD meet at X. Prove that (XA)(XB) = (XC)(XD). (No citing the Power of aPoint Theorem! You’re being asked here to prove it!)

Solution. We take the case where X lies inside the circle. (The case where X lies outside the circleis handled similarly.)

A

B

C

D

X

Both ∠BAC and ∠BDC subtend arc BC, so ∠BAC = ∠BDC. Both ∠ACD and ∠ABD subtendarc AD, so ∠ACD = ∠ABD. Hence, triangles XAC and XDB are similar, which means XA/XD =XC/XB. This implies XA ·XB = XC ·XD.

7. Point P is outside a circle and A is on the circle such that PA is tangent to the circle. A line throughP meets the circle at B and C. Prove that PA2 = (PB)(PC). Use this to show that if a second linethrough P meets the circle at D and E, then PA2 = (PD)(PE). (Again, you can’t just cite Power ofa Point!)

Solution. Since AP is tangent to the circle at A, ∠PAC = ∠ABC = ∠PBA. Hence, triangles PACand PBA are similar. It follows that PA/PB = PC/PA, so PA2 = PB ·PC. By the same argument,PA2 = PD · PE.

A

B

C

DEP

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

8. In the diagram below, we have BP = 8, AB = 10, CD = 7, and ∠APC = 60◦. Find the area of thecircle. Source: AHSME

P

DC

B

A

Solution. Let x = PC. Then by power of a point on P , PA · PB = PC · PD, or 18 · 8 = x(x + 7),which simplifies to x2 + 7x− 144 = 0. This equation factors as (x− 9)(x+ 16) = 0, so x = 9.

A

B

CD

P

10

8

79

Since PC/PA = 9/18 = 1/2 and ∠APC = 60◦, triangle APC is a 30◦-60◦-90◦ triangle. In particular,AC = CP

√3 = 9

√3, and ∠ACP = 90◦, so ∠ACD = 90◦. Then by Pythagoras on right triangle

ACD, AD2 = AC2 + CD2 = (9√

3)2 + 72 = 292. Since ∠ACD = 90◦, AD is a diameter of the circle,so the area of the circle is

π ·(AD

2

)2

=AD2

4· π = 73π.

9. Point O is the center of the circle, AB ⊥ BC, AP = AD, and AB has length twice the radius of thecircle. Prove that AP 2 = (PB)(AB).

D

B

A C

EP

O

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

Solution. Let r be the radius of the circle. Since AB is perpendicular to BC, and BC is a diameter ofthe circle, AB is tangent to the circle at B. Then by power of a point on A, AD ·AE = AB2 = (2r)2 =4r2. But AE = AD+DE = AD+2r, so AD(AD+2r) = 4r2. Expanding, we get AD2 +2rAD = 4r2,so

AD2 = 4r2 − 2rAD = (2r −AD)(2r).

Since AD = AP , AD2 = AP 2. Also, PB = AB −AP = 2r −AD, and AB = 2r, so AD2 = PB ·AB,as desired.

10. (This one can be particularly useful.) Show that if P is outside circle �O, and that the radius of �Ois r, then the power of point P is OP 2 − r2. What if P is inside the circle?

Solution. Let the line through O and P intersect the circle at A and B. Then by power of a point onP , PA · PB = (OP +AO)(OP −BO) = (OP + r)(OP − r) = OP 2 − r2.

A

B

O

P

Now, let P be inside the circle. As above, let the line through O and P intersect the circle at A andB. Then PA · PB = (AO +OP )(OB −OP ) = (r +OP )(r −OP ) = r2 −OP 2.

A

B

OP

However, it is more conventional to define the power of a point P as OP 2 − r2. Note that under thisdefinition, the power is positive if P lies outside the circle, and is negative if P lies inside the circle.(This simplifies the formula for the power of a point. The other idea behind this definition is the useof directed line segments. For example, in the first diagram, PA and PB point in the same direction,so their product is considered positive. However, in the second diagram, PA and PB point in oppositedirections, so their product is considered negative.)

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

11. Describe the radical axis of two intersecting circles.

Solution. Consider one of the points of intersection. Since this point lies on both circles, the power ofthis point with respect to both circles is 0, which means that it lies on the radical axis of both circles.The same holds for the other intersection.

We know that the radical axis is a line, so the radical axis of two intersecting circles is the line passingthrough their points of intersection. Note that if the two circles are tangent, then the radical axis istheir common internal tangent.

12. Circles C1 and C2 meet at A and B. Circles C2 and C3 meet at C and D. Circles C1 and C3 meet at Eand F . Show that

←→AB,

←→CD, and

←→EF are concurrent.

Solution. Let R be the intersection of AB and CD.

C1 C2

C3

A

B C

D

E

FR

By Exercise 11, AB is the radical axis of C1 and C2, so R has the same power with respect to circlesC1 and C2. Similarly, CD is the radical axis of C2 and C3, so R has the same power with respect tocircles C2 and C3. Hence, R has the same power with respect to circles C1 and C3.

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

Therefore, R lies on CD, the radical axis of C1 and C3. This proves that AB, CD, and EF areconcurrent at R.

(We can use the same argument to prove that given any three circles, the radical axes of the circlestaken pairwise are concurrent. The point of concurrence is called the radical center of the three circles.)

13. Altitudes AD, BE, and CF of acute triangle 4ABC intersect at H. Show that ∠BFD = ∠ACB.

Solution. Since ∠ADC = ∠AFC = 90◦, quadrilateral ACDF is cyclic. (In fact, D and F lie on thecircle with diameter AC.) Hence, ∠BFD = 180◦ − ∠AFD = ∠ACD = ∠ACB.

A

B CD

E

F

14. Points A, C, D, and B are in that order on a circle with center O such that AB is a diameter of thecircle and ∠DOB = 50◦. Point P is on AB such that ∠PCO = ∠PDO = 10◦. Find the measure indegrees of arc CD.

Solution. Since ∠PCO = ∠PDO = 10◦, quadrilateral OPCD is cyclic. Then ∠PCD = 180◦ −∠POD = ∠DOB = 50◦, so ∠OCD = ∠PCD − ∠PCO = 50◦ − 10◦ = 40◦.

A B

C

D

OP

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

Since OC = OD, triangle OCD is isosceles, so ∠ODC = ∠OCD = 40◦. Then ∠COD = 180◦ −∠ODC − ∠OCD = 180◦ − 40◦ − 40◦ = 100◦, so arc CD is 100◦.

15. Prove that each of the methods above does indeed show that a quadrilateral is cyclic.

• If ∠ACB = ∠ADB in convex quadrilateral ABCD, then ABCD is cyclic.

Let X be the intersection of diagonals AC and BD. Since ∠BCX = ∠ADX and ∠BXC = ∠AXD,triangles BCX and ADX are similar. Hence, BX/AX = CX/DX, so BX ·DX = AX · CX.

A

B

C

D

X

A

B

C

D′

X

Let the circumcircle of triangle ABC intersect BD at D′. Then by power of a point on X, BX ·D′X =AX · CX. Hence, points D and D′ coincide, which means quadrilateral ABCD is cyclic.

• If ∠BAD + ∠BCD = 180◦ in convex quadrilateral ABCD, then ABCD is cyclic.

Let X be the intersection of AB and CD. Note that ∠XAD = 180◦ − ∠BAD = ∠BCD = ∠BCX.Also, ∠AXD = ∠CXB. Hence, triangles AXD and CXB are similar. It follows that AX/CX =DX/BX, so AX ·BX = CX ·DX.

A

B

CD

X

A

B

CD′

X

Let the circumcircle of triangle ABC intersect CX at D′. Then by power of a point on X, AX ·BX =CX ·D′X. Then points D and D′ coincide, so quadrilateral ABCD is cyclic.

• If P is the intersection of AB and CD, and PA·PB = PC ·PD, then ABCD is a cyclic quadrilateral.

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

From PA · PB = PC · PD, we get PA/PC = PD/PB. Since ∠APD = ∠DPB, triangles APD andCPB are similar. Then ∠BCD = ∠BCP = ∠DAP = 180◦ − ∠BAD, so quadrilateral ABCD iscyclic.

A

B

CD

P

16. Prove Ptolemy’s Theorem. This is pretty tough to do if you’ve never seen the proof before, so I’ll giveyou the first step. Ptolemy’s Theorem involves a bunch of side lengths and a circle, so we might thinkof Power of a Point. Unfortunately, most of the side lengths are chords of the circle and are not wellsituated to use Power of a Point. However, circles give us equal angles, which can help find similartriangles. So, here’s the first step: take a point M on diagonal BD such that ∠ACB = ∠MCD. Now,find some similar triangles. Then mark equal angles and find more similar triangles.

Solution. Take point M on BD so that ∠ACB = ∠MCD. Also, ∠BAC = ∠BDC = ∠MDC, sotriangles ABC and DMC are similar. Hence,

DM

AB=CD

AC,

so DM = AB · CD/AC.

A B

C

D

M

Subtracting equal angles ∠ACB and ∠MCD from ∠BCD, we find ∠ACD = ∠BCM . Also, ∠DAC =∠DBC = ∠MBC, so triangles ACD and BCM are similar. Hence,

BM

AD=BC

AC,

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Art of Problem Solving

WOOT 2012–13

Similar Triangles/Power of a Point

so BM = AD ·BC/AC.

From our two equations,

AB · CDAC

+AD ·BCAC

= BM +DM = BD,

so AB · CD +AD ·BC = AC ·BD.

17. Let ABC be an equilateral triangle and let P be a point on its circumference such that PC is greaterthan both PA and PB. Prove that PA+ PB = PC.

Solution. For PC to be greater than both PA and PB, P must lie on arc AB.

A

B C

P

By Ptolemy’s theorem on quadrilateral APBC,

PA ·BC + PB ·AC = PC ·AB.

Since AB = AC = BC, it follows that PA+ PB = PC.

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