What is the measure of the $\angle AHC$ in the figure below? For reference: The circumference ex-inscribed in a $\triangle ABC$ determines the tangency points F and G in BC and the prolongation of AB, respectively, the prolongation of GF crosses AO at point H, where O is the center of the ex-inscribed circle. Calculate  $\angle AHC$
My progress...I managed to draw

relationships I found:
$\measuredangle MAO = \measuredangle OAG\\
\measuredangle GOB=\measuredangle BOA\\
\measuredangle CBO = \measuredangle OBG\\
FBGO(cyclic)\\
BG=BF, AG = AM $
 A: If you call $\angle ACB = 2\gamma$ then because $CO$ is the exterior angle bisector, since $O$ is the excentre, we have $\angle OCM = 90 - \gamma$. Hence $\angle COM = \gamma$.
This means that $\angle AOC = \angle AOM - \gamma = 90 - \alpha - \gamma = 90 - \beta$ (because $90 - \alpha - \gamma$ is the measure of half the interior angle at $B$ which is $90 - \beta$).
On the other hand $\angle HFC = 90 + \angle GFO = 90 + \beta$, since $F$ is the point of tangency, and $GBFO$ is cyclical. We conclude $\angle CFH + \angle COH = 90 + \beta + 90 - \beta = 180$. Thus $CFHO$ is cyclical.
From this $\angle CHO = 90$ (and hence also $AHC$) since its opposite to the straight angle at $M$ in this cyclic quadrilateral.
A: 
I refer to internal angles of $\triangle ABC$ as $\angle A, \angle B, \angle C$.
As $\angle GBF = 180^\circ - \angle B$ and $BG = BF$,
$\angle AGH = \frac{\angle B}{2}$
So, $\angle OHG = \angle AGH + \angle GAH = \frac{\angle A + \angle B}{2} = 90^0 - \frac{\angle C}{2}$
Next, $\triangle AOG \cong \triangle AOM$ (by S-A-S). As $OA$ is angle bisector of $\angle GOM$ and $OG = OM$, every point on $OA$ is equidistant from points $G$ and $M$ and follows that $\triangle OHG \cong \triangle OHM$.
So, $\angle OHM = \angle OHG = 90^\circ - \frac{\angle C}{2} = \angle OCM$
As segment $OM$ subtends same angle at points $C$ and $H$, $OHCM$ is cyclic and $\angle AHC = \angle OMC = 90^\circ$
