62. In an isosceles right triangle $ABC$, $D$ is a point on the hypotenuse $AB$, and $CD=5, BD-AD=6$, then the area of $\triangle ACD$ is $\qquad$.
Answer: 2
Solution: As shown in the figure, draw $C E \perp A B$ at $E$, let $A D=x$, then
$$
B D=x+6, A B=A D+B D=x+x+6=2 x+6 \text {, }
$$

Therefore, $A E=\frac{1}{2} A B=\frac{1}{2} \times(2 x+6)=x+3, D E=A E-A D=x+3-x=3$.

In the right triangle $\triangle C D E$,
$$
C E^{2}=C D^{2}-D E^{2}=5^{2}-3^{2}=16,
$$

Thus,
$$
C E=4 \text {. }
$$

Therefore,
$$
A E=C E=4 .
$$
$$
\begin{array}{l}
A D=A E-E D=4-3=1, \\
S_{\triangle A C D}=\frac{1}{2} A D \cdot C E=\frac{1}{2} \times 1 \times 4=2 .
\end{array}
$$

Hence, the answer is 2.