This article is from the Puzzles FAQ, by Chris Cole firstname.lastname@example.org and Matthew Daly email@example.com with numerous contributions by others.
Prove if two angle bisectors of a triangle are equal, then the triangle is
isosceles (more specifically, the sides opposite to the two angles
being bisected are equal).
PROVE: <ABC = <BCA (i.e. triangle ABC is an isosceles triangle)
A / \ / \ D E XP normal to AB / \ / \ XQ normal to AC P /----X----\ Q / / \ \ / / \ \ / / \ \ B/_______________\C
Let's assume XD > XE. Then ang(PDX) < ang(QEX) Now considering triangles BXD and CXE, the last condition requires that ang(DBX) > ang(ECX) OR ang(XBC) > ang(XCB) OR XC > XB Thus our assumption leads to : XC + XD > XE + XB OR CD > BE which is a contradiction.