# 1961 AHSME Problems/Problem 10

## Problem

Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$, and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is:

$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$

## Solution

$[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label("B",(0,0),SW); dot((50,0)); label("C",(50,0),SE); dot((25,43.301)); label("A",(25,43.301),N); dot((25,0)); label("D",(25,0),S); dot((25,21.651)); label("E",(25,21.651),E); draw((25,21.651)--(0,0)); label("12",(10,25)); label("6",(12.5,-5)); label("6",(37.5,-5)); label("12",(40,25)); draw((25,3)--(28,3)--(28,0)); label("3\sqrt{3}",(30.5,11)); [/asy]$ Note that $\triangle ABC$ is an equilateral triangle. From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$. That means $DE = 3\sqrt{3}$. Using the Pythagorean Theorem again, $BE = \sqrt{63}$, which is answer choice $\boxed{\textbf{(D)}}$.