geometry
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 $
If I understand correctly, $E$ is on the segment $\overline{AB}$ and the line $\overleftrightarrow{CE}$ intersects the arc $\overset{\large\frown}{AB}$ at the midpoint of the arc; $F$ is on the segment $\overline{BC}$ and the line $\overleftrightarrow{AF}$ intersects the arc $\overset{\large\frown}{BC}$ at the midpoint of the arc.
Therefore $\overrightarrow{CE}$ is the bisector of the angle $\angle ACB$ and $\overrightarrow{AF}$ is the bisector of the angle $\angle BAC.$ Therefore $\overrightarrow{CE}$ and $\overrightarrow{AF}$ intersect at $D,$ the center of the inscribed circle, as shown in your figure.
Because $\overrightarrow{CE}$ and $\overrightarrow{AF}$ are angle bisectors, because $\angle ABC$ is a right angle, and because $\overline{EH}$ and $\overline{FG}$ are perpendicular to $\overline{AC},$ it follows that $\triangle ABF \cong \triangle AGF$ and $\triangle CBE \cong \triangle CHE.$
The relationships of the segments $\overline{EH}$ and $\overline{FG}$ to the inscribed circle should then be obvious (namely, they are exactly as they appear in your figure). Then the relationship of the distance $GH$ to the radius of the inscribed circle is easily seen.
$\angle BAC + \angle FAE = 65^0$
$\angle PAQ = \frac{\angle POQ}{2} = 90^\circ - \frac{x}{2}$
$\angle BAF + \angle BDF = 180^\circ$
$\implies 65^\circ + 90^\circ - \frac{x}{2} + x = 180^\circ$
$\implies x = 50^\circ$
Best Answer
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.