Given a convex quadrilateral $ KLMN $, in which $ \angle NKL = {{90} ^ {\circ}} $. Let $ P $ be the midpoint of the segment $ LM $. It turns out that $ \angle KNL = \angle MKP $. Prove that $ \angle KNM = \angle LKP $.
Source: 2016 Kyiv TST1 9.3 for Ukraine MO
Tags: equal angles, geometry
Given a convex quadrilateral $ KLMN $, in which $ \angle NKL = {{90} ^ {\circ}} $. Let $ P $ be the midpoint of the segment $ LM $. It turns out that $ \angle KNL = \angle MKP $. Prove that $ \angle KNM = \angle LKP $.