How to prove that a quadrilateral with a circle inscribed inside it is cyclic? The question is given as follows:
In the diagram below, $BF \perp HD$. Prove that $ACEG$ is a cyclic quadrilateral. 
In class, we were told to introduce the origin $O$ and draw radii $OB, OH, OF,$ and $OD$. We then noticed that $\angle FOD = 180°- \angle FED$ and $\angle HOB = 180° -\angle HAB $, were told after that to use Thales' Theorem twice. At this point, I don't know where to use Thales' Theorem, or where to go with the problem.
 A: Construction: Extend DO and HO to cut the circle at M and N respectively.

By considering the pink quadrilateral, we have the red marked angles are equal.
Similarly, the green marked angles are equal.
I forget the name of the following theorem (probably is called extended inscribed angle theorem). That fact is:-The size of the angle at X (HXB) is proportional to (arc HB + arc FD).
Since the angle at $X = 90^0$, then the proportional arc length is half the circumference = (arc MF + arc FD).
This means arc HB = arc MF. Then, $\angle purple = \angle red$.
Result follows because $\angle green + \angle red =\angle green + \angle purple = 180^0$.
Added: Explaining "the theorem".

Let X be the point of intersection of chords AB and CD. It should be clear that:-
$\angle 3 = \angle 2 + \angle 1 = \angle2 + \angle 1’$
$\angle 2$ is proportional to the blue arc BD and $\angle 1’$ is proportional to the green arc AC. Hence, we get the required result.
A: Here is another solution. Extend $AC$ and $GE$ until they intersect at a point $P$. Then since $AC$ and $GE$ are tangent to the circle at points $B$ and $F$ respectively, $PB = PF$ and therefore triangle $PBF$ is isosceles which means that $$\angle \, FBA = \angle \, FBP = \angle \, BFP = \angle\, BFG = \alpha$$ In the case when $AC$ and $GE$ do not intersect, they are parallel and then $$\angle \, FBA = \angle \, FBP = 90^{\circ} = \angle \, BFP = \angle\, BFG = \alpha$$
Analogously, we can conclude that 
$\angle \, AHD = \angle \, CDH = \beta$. 
What follows is just angle chasing. Let $K$ be the intersection point of $HD$ and $BF$. In quad $ABKH$ the following angle identity holds $$360^{\circ} = \angle \, A  + \angle \, ABK + \angle \, AHK + \angle \, BKH = \angle \, A + \alpha + \beta + 90^{\circ}$$ so 
$$\angle \, A = 270^{\circ} - \alpha - \beta$$
 In quad $DEFK$ the following angle identity holds $$360^{\circ} = \angle \, E  + \angle \, EFK + \angle \, EDK + \angle \, DKF = \angle \, A + (180^{\circ} - \alpha) + (180^{\circ} - \beta) + 90^{\circ}$$ so 
$$\angle \, E = \alpha + \beta - 90^{\circ}$$
Finally calculate $$\angle \, A + \angle \, E = 270^{\circ} - \alpha - \beta \, + \,  \alpha + \beta - 90^{\circ} = 180^{\circ}$$ which holds exactly when the quadrilateral $ACEG$ is cyclic.
