In triangle ABC, the perpendiculars from A to the internal bisectors of angles B and C meet those bisectors at X and Y. Prove that XY is parallel to BC.

Extending AX to its point of intersection with BC (let's call this point D) and AY to its point of intersection with BC (let's call this point E) creates two isoceles triangles ADB and AEC with X the midpoint of AD and Y the midpoint of AE. AXY is then similar to ADE with XY being parallel to DE which is on BC.

Posted by Eric on 2006-11-27 12:23:47