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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 $\angle {GAC}$. enter image description here

I have found that: $\angle E=\frac{\angle B}{2}$ and $\angle ACG=90-\frac{\angle C}{2}$ but can't go further...

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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 $\angle GAC=\frac{\gamma}{2}$.

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    Presumably you meant: $$\angle BED = \angle BDE = \angle GDC = \frac{\beta}{2}$$ – 2017-01-30
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    Edited, thanks! – 2017-01-30
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    Why is $\angle AOE=\frac{\alpha}{2}$? Isn't it $\angle OAE=\frac{\alpha}{2}$? – 2017-01-30
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    A typo, again... – 2017-01-30
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    How do you infer that quadrilateral $AGOE$ is cyclic? – 2017-01-30
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    $\angle EAO=\angle EGO=\frac{\alpha}{2}$. – 2017-01-30
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    Got it! Very nice. – 2017-01-30
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    @Sz_Z What is $\gamma$? – 2017-01-30
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    The angles of the triangle $ABC$ are $\alpha$, $\beta$, $\gamma$, as usual. – 2017-01-30