3. In $\triangle A B C$, it is known that $\overrightarrow{B C} \cdot \overrightarrow{B A}=\overrightarrow{C B} \cdot \overrightarrow{C A}$. Then $\triangle A B C$ is ( ).
(A) Isosceles triangle
(B) Right triangle
(C) Isosceles right triangle
(D) None of the above
3. A.

From the problem, we know that $\overrightarrow{B C} \cdot(\overrightarrow{A B}+\overrightarrow{A C})=0$.
If we denote the midpoint of side $B C$ as $D$, then $\overrightarrow{A B}+\overrightarrow{A C}=2 \overrightarrow{A D}$. Therefore, $\overrightarrow{B C} \cdot \overrightarrow{A D}=0$. This indicates that $B C \perp A D$. Hence, $\triangle A B C$ is an isosceles triangle.