Let $ABC$ is isosceles triangle. $AB=AC$. Point $P$ such that $\angle BPC=2\angle BAC$. $PK$ is bisector of $\angle BPD$ and $AK \perp PK$. Prove that $$2AK=BP+PC$$
Let point $A'$ and $B'$ such that $AK=KA'$ and $PB=PB'$. Then need prove that $AA'=B'C$.
This is equivalent to proving that the $ACA'B'$ is an isosceles trapezoid.