Problem

Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2018

Tags: geometry, Triangle, isosceles



In triangle $ABC$ given is point $P$ such that $\angle ACP = \angle ABP = 10^{\circ}$, $\angle CAP = 20^{\circ}$ and $\angle BAP = 30^{\circ}$. Prove that $AC=BC$