In a triangle $ABC$, $AB > AC$, the external bisector of angle $A$ meets the circumcircle of triangle $ABC$ at $E$, and $F$ is the foot of the perpendicular from $E$ onto $AB$. Prove that $2AF = AB - AC$
Source: 2000 Singapore TST 1.1
Tags: geometry, circumcircle, angle bisector
In a triangle $ABC$, $AB > AC$, the external bisector of angle $A$ meets the circumcircle of triangle $ABC$ at $E$, and $F$ is the foot of the perpendicular from $E$ onto $AB$. Prove that $2AF = AB - AC$