2001 AIME II Problem 13

Question

In quadrilateral

ABCD,

\angle BAD \cong \angle ADC

and

\angle ABD \cong \angle BCD,

AB = 8,

BD = 10,

and

BC = 6.

The length

CD

may be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 69

Concepts:similarity isosceles triangle angle chasing

Difficulty rating: 2990

Small hint:

Extend

\overline{AB}

and

\overline{DC}

to meet at

P;

the equal angles at

A

and

D

force

PA = PD

Big hint:

Angles

PBD

and

PCB

are supplements of the given equal angles, so triangles

PCB

and

PBD

are similar

Solution:

Extend

\overline{AB}

beyond

B

and

\overline{DC}

beyond

C

to meet at

P.

Since

\angle PAD = \angle PDA,

triangle

APD

is isosceles with

PA = PD.

Also

\angle PBD = 180^\circ - \angle ABD

and

\angle PCB = 180^\circ - \angle BCD,

so

\angle PBD = \angle PCB.

Triangles

PCB

and

PBD

share angle

P

and have

\angle PCB = \angle PBD,

so they are similar, giving

\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}.

Since

PB = PA - 8 = PD - 8,

the middle ratio reads

\frac{PD - 8}{PD} = \frac{3}{5},

so

PD = 20

and

PB = 12.

Then

PC = \frac{3}{5} \cdot 12 = \frac{36}{5}.

Finally

CD = PD - PC

= 20 - \frac{36}{5}

= \frac{64}{5},

which is in lowest terms, so

m + n = 64 + 5 = 69.