9. $\triangle A B C$ is isosceles with $A B=A C . P$ is a point inside $\triangle A B C$ such that $\angle B C P=$ $30^{\circ}, \angle A P B=150^{\circ}$ and $\angle C A P=39^{\circ}$. Find $\angle B A P$.
(1 mark)
$A B C$ 是等腰三角形, 其中 $A B=A C \circ P$ 是 $\triangle A B C$ 内的一點, 使得 $\angle B C P=$ $30^{\circ} 、 \angle A P B=150^{\circ}$ 且 $\angle C A P=39^{\circ}$ 。求 $\angle B A P$ 。
9. $13^{\circ}$
9. Let $D$ be the circumcentre of $\triangle B C P$. We have $\angle P D B=2 \angle P C B=60^{\circ}$ and $D B=D P$, which means $\triangle B D P$ is equilateral and so $\angle A P D=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}=\angle A P B$. Together with $P B=P D$, we have $\triangle A P B \cong \triangle A P D$.
Next, from $A D=A B=A C$ and $D B=D C$, we get that $\triangle A B D \cong \triangle A C D$. From these, we have $\angle B A P=\angle D A P$ and $\angle B A D=\angle C A D$. It follows that $\angle B A P=\frac{1}{3} \angle C A P=13^{\circ}$.