12 Let $E$ be the intersection of the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$. Given: $AC=BC, AD=5, BE=12, DE=3$, find $\angle BCD$.
Let $O$ be the intersection of the perpendicular bisector of $C D$ and $B D$, then
$$
\angle C O D=180^{\circ}-2 \angle O D C=180^{\circ}-2 \angle B A C=\angle A C B=\angle A D O \text {. }
$$

This implies $A D / / C O \Rightarrow \frac{O E}{3}=\frac{C O}{A D}=\frac{O E+3}{5} \Rightarrow O E=\frac{9}{2}$. Thus, $O$ is the midpoint of $B D$, hence $\angle B C D=90^{\circ}$.