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

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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''.

Contributed by: Jay Warendorff (March 2011)

Open content licensed under CC BY-NC-SA

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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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