21. The diagram below shows a triangle $A B C$ in which $\angle A=60^{\circ}, B P$ and $B E$ trisect $\angle A B C$; and $C P$ and $C E$ trisect $\angle A C B$. Let the angle $\angle B P E$ be $x^{\circ}$. Find $x$.
21. Ans: 50 .
$$
\angle B P C=180^{\circ}-(\angle P B C+\angle P C B)=180^{\circ}-\frac{2}{3}(\angle A B C+\angle A C B)=180^{\circ}-\frac{2}{3} 120^{\circ}=100^{\circ} .
$$

Observe that $P E$ bisects $\angle B P C$, the result follows.