Problem

Source: Serbia IMO TST 2019 P2

Tags: geometry



Given triangle $\triangle ABC $ with $AC\neq BC $,and let $D $ be a point inside triangle such that $\measuredangle ADB=90^{\circ} + \frac {1}{2}\measuredangle ACB $.Tangents from $C $ to the circumcircles of $\triangle ABC $ and $\triangle ADC $ intersect $AB $ and $AD $ at $P $ and $Q $ , respectively.Prove that $PQ $ bisects the angle $\measuredangle BPC $.