2. In a convex quadrilateral $ABCD$, $AB>AD$, $BC=CD$, $AC$ bisects $\angle BAD$, $\angle BAD=25^{\circ}$. Then the degree measure of $\angle BCD$ is ( ).
(A) $160^{\circ}$
(B) $155^{\circ}$
(C) $150^{\circ}$
(D) $145^{\circ}$
2. B.

As shown in Figure 3, take point $D_{1}$ on $AB$ such that $A D_{1}=A D$, and connect $C D_{1}$.
Then point $D$ and $D_{1}$ are symmetric with respect to $A C$.
$$
\begin{array}{l}
\Rightarrow \angle A D C=\angle A D_{1} C, B C=C D=C D_{1} \\
\Rightarrow \angle A D C+\angle A B C=180^{\circ} .
\end{array}
$$

From the fact that the sum of the interior angles of a convex quadrilateral is $360^{\circ}$, we know
$$
\angle B C D=180^{\circ}-\angle B A D=155^{\circ} \text {. }
$$