In an isoceles triangle ABC the angle ABC equals 135 degrees. The point F lies on AC. The angle ABF equals 90 degrees. AG is the bisector of the angle BAC, where G is the intersection point of BC and the bisector. How can I find the value of the angle GFC?
It's quite easy to find the angle A, but what shall I do next? I tried finding many angles, finding sums of the triangle's angles to find the value of GFC, but it was worthless.
Well, it's simple. If you calculate the various angles around vertex $B$ you realize that $BC$ is the exterior angle bisector of triangle $ABF$ through vertex $B$. Also, $AG$ is the angle bisector of $\angle\, BAC = \angle \, BAF$. Since the exterior angle bisectors of vertices $B$ and $F$ of triangle $ABF$ and the interior angle bisector of its vertex $A$ must intersect at a common point, that common point is $G = AG \cap BC$. Hence, $FG$ is the exterior angle bisector of $ABF$ through $F$. Thus, $$\angle \, GFC = \frac{1}{2} \,\angle BFC = \frac{1}{2} \,(\angle \, BAF + \angle \, ABF) = \frac{1}{2} \left(\frac{45^{\circ}}{2} + 90^{\circ}\right)$$