jmc

Let $BE=x$ and $BC=y$. Since $AF\parallel BC$, by AA Similarity, $△AFE\sim △CBE$. 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, $△FDG=△FAB$. 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, $x=\boxed{16}$.