MathNet 2018-part-06 — 181 problems (EN–VI) Question 118

Point $P$ was chosen inside the triangle $ABC$ so that $BC = AP$ and $\angle APC = 180^\circ - \angle ABC$. On side $AB$, there exists point $K$ such that $AK = KB + PC$. Prove $\angle AKC = 90^\circ$.