Perpendiculars from a Point on the Line between the Endpoints of the Angle Bisectors

Let ABC be a triangle. Let AA' and BB' bisect the angles CAB and ABC. Let P be a point on A'B' and PQ be perpendicular to AB. Let PB'' and PA'' be perpendicular to AC and BC. Then PQ = PA'' + PB''.



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See problem 1.13 in V. Prasolov, Problems in Plane and Solid Geometry, Vol. 1, Plane Geometry [PDF], (D. Leites, ed. and trans.).
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