13. In parallelogram $A B C D, \angle B A D=76^{\circ}$. Side $A D$ has midpoint $P$, and $\angle P B A=52^{\circ}$. Find $\angle P C D$.
Answer: $38^{\circ}$

Solution: Note that $\angle B P A=180^{\circ}-76^{\circ}-52^{\circ}=52^{\circ}$. Since $\angle P B A=52^{\circ}$, then $\triangle B P A$ is isoceles and $|A B|=$ $|A P|=|P D|$. But $|A B|=|C D|$, so by transitivity $|P D|=|C D|$ and therefore $\triangle P C D$ is also isoceles. Since $\angle C D A=180^{\circ}-76^{\circ}=104^{\circ}$, then $\angle P C D=$ $\frac{180^{\circ}-104^{\circ}}{2}=38^{\circ}$