# 1963 AHSME Problems/Problem 25

## Problem

Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$. The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is: $[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("A", A, SW); label("B", B, S); label("C", C, N); label("D", D, NW); label("E", E, SE); label("F", F, W); //Credit to MSTang for the asymptote[/asy]$ $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

## Solution

Because $ABCD$ is a square, $DC = CB = 16$, $DC \perp DA$, and $CB \perp BA$. Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$, $\angle DCF = \angle BCE$. Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$.

From the congruency, $CF = CE$. Using the area formula for a triangle, $CE = 20$. Finally, by the Pythagorean Theorem, $BE = \sqrt{20^2 - 16^2} = 12$, which is answer choice $\boxed{\textbf{(A)}}$.

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