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:

$[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$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 $\boxed{\textbf{(E)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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1963 AHSME Problems/Problem 38

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