1.2. $A B C$ is a triangle with $\angle A>90^{\circ}$. On the side $B C$, two distinct points $P$ and $Q$ are chosen such that $\angle B A P=\angle P A Q$ and $B P \cdot C Q=B C \cdot P Q$. Calculate the size of $\angle P A C$.
1.2. Stewart's theorem. In $\triangle A B C, D$ is a point on $B C$ such that $A D$ bisects $\angle A$. Then $A B: B D=A C: C D$.
1st solution
Applying Stewart's theorem to $\triangle A B Q$, we have $\frac{A B}{A Q}=\frac{B P}{P Q}$.
Given $B P \cdot C Q=B C \cdot P Q$, it follows that $\frac{B C}{C Q}=\frac{A B}{A Q}$.
Now let $R$ be the point on $A C$ such that $Q R$ is parallel to $B A$.
Then $\frac{A B}{R Q}=\frac{B C}{C Q}=\frac{A B}{A Q}$.
Hence $R Q=A Q$ and $\angle Q A R=\angle Q R A$.
Therefore $\angle P A C=\angle P A Q+\angle Q A R=\frac{1}{2}(\angle B A Q+\angle Q A R+\angle Q R A)=\frac{\pi}{2}$.
2nd solution
Since $\frac{C B}{C Q}=\frac{P B}{P Q}=\frac{A B}{A Q}$, by Stewart's theorem, $A C$ is the external angle bisector of $\angle B A Q$. Hence $\angle P A C=\frac{\pi}{2}$.