Let $P$ be the point inside the square $A B C D$ such that $\triangle A B P$ is equilateral. Calculate the angle $\angle C P D$. Explain your answer!
The triangle $D A P$ is isosceles because $A D=A P$ hence $\angle A D P=\angle A P D=\frac{180^{\circ}-\angle D A P}{2}=75^{\circ}$. Hence $\angle P D C=15^{\circ}$. Similarly $\angle D C P=15^{\circ}$ and hence $\angle C P D=180^{\circ}-2 \cdot 15^{\circ}=150^{\circ}$.