4. (12 points) As shown in the figure, in the isosceles right triangle $ABC$, there is a point $D$ on the hypotenuse $AB$. It is known that $CD=5$, and $BD$ is 2 units longer than $AD$. What is the area of triangle $ABC$? $\qquad$ .
【Solution】Solution: Draw $C E \perp A B$ at $E$.
$$
\begin{array}{l}
\because C A=C B, \quad C E \perp A B, \\
\therefore C E=A E=B E, \\
\because B D-A D=2, \\
\therefore B E+D E-(A E-D E)=2, \\
\therefore D E=1,
\end{array}
$$

In Rt $\triangle C D E$, $C E^{2}=C D^{2}-D E^{2}=24$,
$$
\therefore S_{\triangle A B C}=\frac{1}{2} \cdot A B \cdot C E=C E^{2}=24 \text {, }
$$

Therefore, the answer is 24.