1998 AHSME Problems/Problem 28

Solution 2

Let $AC=2$ and $AD=3$ . By the Pythagorean Theorem, $CD=\sqrt{5}$ . Let point $P$ be on segment $CD$ such that $AP$ bisects $\angle CAD$ . Thus, angles $CAP$ , $PAD$ , and $DAB$ are congruent. Applying the angle bisector theorem on $ACD$ , we get that $CP=\frac{2\sqrt{5}}{5}$ and $PD=\frac{3\sqrt{5}}{5}$ . Pythagorean Theorem gives $AP=\frac{\sqrt{5}\sqrt{24}}{5}$ .

Let $DB=x$ . By the Pythagorean Theorem, $AB=\sqrt{(x+\sqrt{5})^{2}+2^2}$ . Applying the angle bisector theorem again on triangle $APB$ , we have $\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\frac{3\sqrt{5}}{5}}$ The right side simplifies to $\frac{\sqrt{24}}{3}$ . Cross multiplying, squaring, and simplifying, we get a quadratic: $5x^2-6\sqrt{5}x-27=0$ Solving this quadratic and taking the positive root gives $x=\frac{9\sqrt{5}}{5}$ Finally, taking the desired ratio and canceling the roots gives $\frac{CD}{BD}=\frac{5}{9}$ . The answer is $\fbox{(B) 14}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1998 AHSME Problems/Problem 28

Solution 2

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