2001 AIME II Problems/Problem 13

Problem

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 $m + n$ .

Solution

Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$ . Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$ , we know that $\triangle ABD \sim \triangle DCE$ . Hence $\angle ADB = \angle DEC$ , and $\triangle BDE$ is isosceles. Then $BD = BE = 10$ .

$[asy] /* We arbitrarily set AD = x */ real x = 60^.5, anglesize = 28; pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5)); D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D); D(A--MP("E",E)--B,d); D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize)); MP("10",(B+D)/2,SW);MP("8",(A+B)/2,W);MP("6",(B+C)/2,NW); [/asy]$

Using the similarity, we have:

$\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5$

The answer is $m+n = \boxed{069}$ .

Extension: Find $AD$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2001 AIME II Problems/Problem 13

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