Question 86
In a triangle $ABC$, $\angle C = 90^\circ$, $AC \lt BC$. Point $K$ inside $BC$ is such that $CK = CA$, and point $D$ inside $CK$ is such that $\angle DAK = \angle BAK$. $F$ and $P$ are, respectively, the feet of perpendiculars from $D$ to $AB$ and from $A$ to $FK$. Prove that $CP = \frac{1}{2}(AF + FD + DA)$.