Given an acute triangle $ABC$ with circumcenter $O$. The circumcircle of $BCH$ and a circle with diameter of $AC$ intersect at $P (P \neq C)$. A point $Q$ on segment of $PC$ such that $PB = PQ$. Prove that $\angle ABC = \angle AQP$
Problem
Source: Indonesian TST for IMO 2023 stage 1: Test 3 - Geometry
Tags: projective geometry, geometry, Indonesia, Indonesia TST