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

Requires a Wolfram Notebook System
Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.
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
Snapshots
Details
See problem 1.13 in V. Prasolov, Problems in Plane and Solid Geometry, Vol. 1, Plane Geometry [PDF], (D. Leites, ed. and trans.).
Permanent Citation