# 1995 AHSME Problems/Problem 8

## Problem

In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }$

## Solution

$[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6); draw(A--B--E--D--E--B--C--A--B--cycle); label("$$A$$",A,W); label("$$B$$",B,E); label("$$C$$",C,SW); label("$$D$$",D,NE); label("$$E$$",E,S); label("$$6$$",F,W); label("$$4$$",G,NW); [/asy]$

$\triangle BAC$ is a $6-8-10$ right triangle with hypotenuse $AB = 10$.

$\triangle BDE$ is similar to $\triangle BAC$ by angle-angle similarity since $E=C = 90^\circ$ and $B=B$, and thus $\frac{BD}{BA} = \frac{DE}{AC}$.

Solving the above for $BD$, we get $BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$.