Let $ABC$ be a triangle with $AB = AC$ and let $D$ be the foot of the height drawn from $A$ to $BC$. Let $P$ be a point inside the triangle $ADC$ such that $\angle APB> 90^o$ and $\angle PAD + \angle PBD = \angle PCD$. The $CP$ and $AD$ lines are cut at $Q$ and the $BP$ and $AD$ lines cut into $R$. Let $T$ be a point in segment $AB$ such that $\angle TRB = \angle DQC$ and let S be a point in the extension of the segment $AP$ (on the $P$ side) such that $\angle PSR = 2 \angle PAR$. Prove that $RS = RT$.