2. In pentagon $A B C D E, A B=B C=C D=D E, \angle B=96^{\circ}$ and $\angle C=\angle D=108^{\circ}$. Find $\angle E$.
(1 mark)
在五邊形 $A B C D E$ 中, $A B=B C=C D=D E, \angle B=96^{\circ}$ 且 $\angle C=\angle D=108^{\circ}$ 。
求 $\angle E$ 。
2. $102^{\circ}$
2. Let $P$ be the intersection of $B D$ and $C E$. Then $\angle P B C=\angle P E D=\angle P C D=\angle P D C=$ $\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$, and so $\angle B C P=\angle B P C=$ $72^{\circ}$. Hence $B P=B C=A B$ and $\angle A B P=96^{\circ}-$ $36^{\circ}=60^{\circ}$. It follows that $\triangle A B P$ is equilateral and $A P=P E$. Thus $\angle A P E=108^{\circ}-60^{\circ}=48^{\circ}$ and $\angle A E P=\frac{180^{\circ}-48^{\circ}}{2}=66^{\circ}$, and so $\angle E=$ $\angle A E D=66^{\circ}+36^{\circ}=102^{\circ}$.