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incenter and midpoints, perpendicularity

Source: MEMO 2016 T6

August 25, 2016
geometryincenterincirclemidpointsmidpoint

Problem Statement

Let ABCABC be a triangle for which AB≠ACAB \neq AC. Points KK, LL, MM are the midpoints of the sides BCBC, CACA, ABAB. The incircle of ABCABC with center II is tangent to BCBC in DD. A line passing through the midpoint of IDID perpendicular to IKIK meets the line LMLM in PP.
Prove that ∠PIA=90∘\angle PIA = 90 ^{\circ}.
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