In triangle $ABC$, the difference between angles $B$ and $C$ is equal to $90^o$, and $AL$ is the angle bisector of triangle $ABC$. The bisector of the exterior angle $A$ of the triangle $ABC$ intersects the line $BC$ at the point $F$. Prove that $AL = AF$.
(Alexander Dzyunyak)
Assume angle $CBA$ is $\beta$. Let $PM$ be the perpendicular bisector of $BC$ where $M$ is the middle point of $BC$ and $P$ is on $BA$. Then angle $CAB$ is $180^o - \beta - (\beta + 90^o) = 90^o - 2 \beta$ and angle $CAL$ = $\frac{1}{2} \angle CAB = 45^o - \beta$. In triangle $CLA$, $\angle CLA = 180^o - (\beta + 90^o) - (45^o - \beta) = 45^o$. It is also well know that $AL$ is perpendicular to $FA$ as the interior bisector is perpendicular to the exterior bisector the the same angle in a triangle. We conclude that $LAF$ is isosceles right triangle. Thus $AL = AF$. My sketch is attached.
