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

[asy]
import olympiad;
unitsize(0.8inch);
pair A,B,C,D;
A = (0,1);
B= (1,0);
C = -B;
D = (0,0);
draw(A--B--C--A,linewidth(1));
draw(A--D,linewidth(0.8));
draw(rightanglemark(C,A,B,s=5));
draw(rightanglemark(C,D,A,s=5));
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
[/asy]

Because $\triangle ABC$ is an isosceles right triangle, $\angle ABC = 45^\circ$.  Since $\angle ADB = 90^\circ$, we know that $\angle DAB = 45^\circ$, so $\triangle ABD$ is also a 45-45-90 triangle.  Similarly, $\triangle ACD$ is a 45-45-90 triangle.  Therefore, $DB=DC = DA = 4\sqrt{2}$, so $BC = BD+DC = 8\sqrt{2}$, and  \[[ABC] = \frac{(AD)(BC)}{2} = \frac{(4\sqrt{2})(8\sqrt{2})}{2} = \boxed{32}.\]
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
In a certain isosceles right triangle, the altitude to the hypotenuse has length $4\sqrt{2}$.  What is the area of the triangle?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
32
In a certain isosceles right triangle, the altitude to the hypotenuse has length $4\sqrt{2}$.  What is the area of the triangle?
test