# Difference between revisions of "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 $\fbox{(B) 14}$.