2. As shown in Figure 3, in $\triangle A B C$, $B D$ bisects $\angle A B C$, $C D \perp B D$ at point $D$, $\angle B A C=\angle A C D, B C=$ $5, B A=11$. Then the area of $\triangle A B C$ is
2. $\frac{132}{5}$.

As shown in Figure 8, extend $CD$ to intersect $AB$ at point $E$. It is easy to see that
$\triangle B E D \cong \triangle B C D$.
Therefore, $B E=B C=5$,
$E D=C D$.
Thus, $A E=A B-B E$
$$
=11-5=6 \text {. }
$$

Also, $\angle B A C=\angle A C D$, so $C E=A E=6$.
Therefore, $C D=\frac{1}{2} C E=3$.
By the Pythagorean theorem,
$$
B D=\sqrt{B C^{2}-C D^{2}}=\sqrt{5^{2}-3^{2}}=4 \text {. }
$$

Thus, $S_{\triangle B C D}=\frac{1}{2} B D \cdot C D=\frac{1}{2} \times 4 \times 3=6$.
Therefore, $S_{\triangle B E C}=2 S_{\triangle B C D}=12$.
Then, $S_{\triangle A B C}=\frac{A B}{B E} S_{\triangle B E C}=\frac{11}{5} \times 12=\frac{132}{5}$.