Given an acute-angled $\vartriangle AB$C with altitude $AH$ ( $\angle BAC > 45^o > \angle AB$C). The perpendicular bisector of $AB$ intersects $BC$ at point $D$. Let $K$ be the midpoint of $BF$, where $F$ is the foot of the perpendicular from $C$ on $AD$. Point $H'$ is the symmetric to $H$ wrt $K$. Point $P$ lies on the line $AD$, such that $H'P \perp AB$. Prove that $AK = KP$.