Given an acute-angled △ABC with altitude AH ( ∠BAC>45o>∠ABC). 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⊥AB. Prove that AK=KP.
Problem
Source: IFYM - XI International Festival of Young Mathematicians Sozopol 2022, Theme for 11-12 grade, 4th round p3
Tags: equal segments, geometry