3. In $\triangle A B C$, $\angle C=90^{\circ}, \angle A$ and $\angle B$ are bisected and intersect at point $P$, and $P E \perp A B$ at point $E$. If $B C=2, A C=3$,
then $A E \cdot E B=$ $\qquad$
3. 3

Solution: Draw $P D \perp A C$ at $D$, $P F \perp B C$ at $F$, then:
$$
\begin{aligned}
C D & =C F \\
& =\frac{1}{2}(A C+B C-A B) \\
& =\frac{1}{2}(5-\sqrt{13}), \\
A D & =A C-C D \\
& =\frac{1}{2}(\sqrt{13}+1), \\
B F & =B C-C F=\frac{1}{2}(\sqrt{13}-1), \\
\therefore \quad & A E \cdot E B=A D \cdot B F=3 .
\end{aligned}
$$