Geometry problem #2

In the animation, every moving triangle preserves its area, because the varying vertex moves parallel to the opposite highlighted base. Hence, the area of $\triangle AFH$ equals half the area of the square $BCED$ . But, $\left[ BCED \right]=\dfrac{1}{2}B{{E}^{2}}$ . Hence, $\left[ AFH \right]=\frac{1}{2}\left[ BCED \right]=\frac{1}{2}\times \frac{1}{2}\times {{\left( 8\sqrt{2} \right)}^{2}}=\boxed{32}$