# 1963 AHSME Problems/Problem 38

## Problem

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals: $[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C); pair E = intersectionpoints(A--C, B--F); draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("A", A, SW); label("B", B, SE); label("C", C, NE); label("D", D, NW); label("F", F, N); label("G", G, NE); label("E", E, SE); //Credit to MSTang for the asymptote[/asy]$ $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

## Solution

Let $BE = x$ and $BC = y$. Since $AF \parallel BC$, by AA Similarity, $\triangle AFE \sim \triangle CBE$. That means $\frac{AF}{CB} = \frac{FE}{BE}$. Substituting in values results in $$\frac{AF}{y} = \frac{32}{x}$$ Thus, $AF = \frac{32y}{x}$, so $FD = \frac{32y - xy}{x}$.

In addition, $DC \parallel AB$, so by AA Similarity, $\triangle FDG = \triangle FAB$. That means $$\frac{\frac{32y-xy}{x}}{\frac{32y}{x}} = \frac{24}{x+32}$$ Cross multiply to get $$\frac{y(32-x)}{x} (x+32) = \frac{32y}{x} \cdot 24$$ Since $x \ne 0$ and $y \ne 0$, $$(32-x)(32+x) = 32 \cdot 24$$ $$32^2 - x^2 = 32 \cdot 24$$ $$32 \cdot 8 = x^2$$ Thus, $x = 16$, which is answer choice $\boxed{\textbf{(E)}}$.

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