Question 12
In an isosceles triangle $ABC$ with $AB = AC$, let $D$ be the foot of the perpendicular of $A$ and $P$ be an interior point of the triangle $ADC$ such that $\angle APB \gt 90^\circ$ and $\angle PBD + \angle PAD = \angle PCB$. Let $Q$ be the intersection of the lines $CP$ and $AD$, and $R$ be the intersection of the lines $BP$ and $AD$. Let $T$ be a point on $[AB]$ and $S$ be a point on the ray $[AP]$ not belonging to $[AP]$ such that $\angle TRB = \angle DQC$ and $\angle PSR = 2\angle PAR$. Show that $RS = RT$.