4. Let points $C$ and $D$ be on the semicircle with diameter $AB$, and line segments $AC$ and $BD$ intersect at point $E$. Given $AB=10$, $AC=8$, and $BD=5\sqrt{2}$. Then the area of $\triangle ABE$ is

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4. $\frac{150}{7}$.

As shown in Figure 2, connect $A D$ and $B C$. Draw $E M \perp A B$ at point $M$.
From the given information, we have
$$
B C=6 \text{. }
$$

Since $A B$ is the diameter of the semicircle, then
Rt $\triangle A C B \backsim \text{Rt} \triangle A M E$,
Rt $\triangle B M E \backsim \text{Rt} \triangle B D A$
$\Rightarrow \frac{A M}{A C}=\frac{E M}{B C}, \frac{E M}{D A}=\frac{B M}{B D}$
$\Rightarrow E M=\frac{3}{4} A M, E M=B M$.
Since $A B=10$, we have $E M=\frac{30}{7}$.
Thus, $S_{\triangle A B E}=\frac{1}{2} A B \cdot E M=\frac{150}{7}$.