In triangle $ABC$ the excircle of $A$ is tangent to sides $BC$ at $D$ and to extensions of $AB$ and $AC$ at points $E$ and $F$ respectively.If $ED$ intersects outer angle bisector of $C$ at $G$ find $\angle {GAC}$.
I have found that: $\angle E=\frac{\angle B}{2}$ and $\angle ACG=90-\frac{\angle C}{2}$ but can't go further...
Let the center of the excircle be $O$. Then $\angle BED = \angle BDE = \angle GDC = \frac{\beta}{2}$, so from the triangle $CDG$ - using that $\angle DCG=\gamma+90^{\circ}-\frac{\gamma}{2}=90^{\circ}+\frac{\gamma}{2}$ - $$\angle DGC =180^{\circ}-\frac{\beta}{2}-(90^{\circ}+\frac{\gamma}{2})=\frac{\alpha}{2}.$$ Also, $\angle EAO=\frac{\alpha}{2}$. Thus the quadrilateral $AGOE$ is cyclic. Since $\angle AEO$ is a right angle, also $\angle AGC$ is a right angle. So from the right triangle $AGC$ we get that $\angle GAC=\frac{\gamma}{2}$.