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 $\boxed{\textbf{(E)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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