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2001 AIME II Problems/Problem 13

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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.

/* We arbitrarily set AD = x */real x = 60^.5, anglesize = 28;pointpen = black; pathpen = black+linewidth(0.7); pen d = linet...

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.

See also

2001 AIME II (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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