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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 $$2AK=BP+PC$$
My attempts:
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.
I dont know how use that $\angle BPC=2\angle BAC$ and $AB=AC$
Because $∠BPC = 2∠BAC$ and $∠DPK = ∠FPK \Rightarrow ∠FDP = ∠DFP$, then $$∠FDP = ∠DFP = \frac{1}{2} ∠BPC = ∠BAC.$$ Therefore,\begin{gather*} ∠ABF = ∠DFP - ∠BAF = ∠BAC - ∠BAF = ∠CAD,\\ ∠BAF = ∠BAC - ∠CAD = ∠FDP - ∠CAD = ∠ACD. \end{gather*} Since $BA = AC$, then$$ △ABF ≌ △CAD. \quad \text{(ASA)} $$ Note that $DP = FP$ and $DK = KF$, thus$$ BP + PC = BF + CD = AD + AF = 2AK. $$
