Let ABCABC be a triangle with AB<AC<BCAB < AC < BC. Let the incentre and incircle of triangle ABCABC be II and ω\omega, respectively. Let XX be the point on line BCBC different from CC such that the line through XX parallel to ACAC is tangent to ω\omega. Similarly, let YY be the point on line BCBC different from BB such that the line through YY parallel to ABAB is tangent to ω\omega. Let AIAI intersect the circumcircle of triangle ABCABC again at PAP \neq A. Let KK and LL be the midpoints of ACAC and ABAB, respectively.

Prove that KIL+YPX=180\angle KIL + \angle YPX = 180^{\circ}.