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

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$