Question 70
An acute triangle $ABC$ is inscribed in circle $\omega$ centered at $O$. Line $BO$ and side $AC$ meet at $B_1$. Line $CO$ and side $AB$ meet at $C_1$. Line $B_1C_1$ meets circle $\omega$ at $P$ and $Q$. If $AP = AQ$, prove that $AB = AC$.