jmc

In isosceles right triangle $△ABC$ below, $\overline{AD}$ is the altitude to the hypotenuse.



Because $△ABC$ is an isosceles right triangle, $\angle ABC=45^{\circ }$. Since $\angle ADB=90^{\circ }$, we know that $\angle DAB=45^{\circ }$, so $△ABD$ is also a 45-45-90 triangle. Similarly, $△ACD$ is a 45-45-90 triangle. Therefore, $DB=DC=DA=4\sqrt{2}$, so $BC=BD+DC=8\sqrt{2}$, and $$[ABC]=\frac{(AD)(BC)}{2}=\frac{(4\sqrt{2})(8\sqrt{2})}{2}=\boxed{32}.$$