High-School level question concerning circle and arcs This question somehow is unsolvable to me. Any idead/hints wil be much appreciated. 

$AB$ is a chord which is cut ny the chords $CD$ and $EC$ in the circle.
Givens:
$\frown{AC} +\frown{BE}=\frown{AD}+\frown{BC}$
$S_{CFG}=S_{CGH}$
I need to show:
$AB \perp HG$
I realized that $CG$ is a median in $\triangle{HFG}$, so I'm trying to prove $CF=CG$ or $CG=GH$ which then will suffice to say that $\triangle{HFG}$ is a right triangle, but am not able to find a way.
 A: Since $$\text{arc}AC+\text{arc}BE=\text{arc}AD+\text{arc}BC,$$
we have
$$\angle{CBA}+\angle{BCE}=\angle{ACD}+\angle{BAC},$$
i.e.
$$\angle{CBG}+\angle{BCG}=\angle{ACF}+\angle{FAC}.$$
Considering $\triangle CBG$ and $\triangle CAF$, we can say
$$\angle{CGF}=\angle{CFG}\Rightarrow CF=CG.$$
Then, as you wrote, since $CF=CH$, we can say $\triangle HFG$ is a right triangle.
A: There is a property of intersecting chords such that the angle between them is equal to the mean of two respective arcs. In the following picture we have $$|\angle AOB| = |\angle COD| = \frac{1}{2}\Big(\overset{\frown}{AB}+\overset{\frown}{CD}\Big).$$ To prove it, just consider $\triangle AOD$ and calculate its angles. Curiously, there are very few references for this fact online, e.g. see here for more info.

This fact makes $\angle CFG$ and $\angle CGF$ equal, and so $\triangle CFG$ is isosceles. In other words, $|CF| = |CG| = |CH|$, and $C$ is a center of circurmcircle of $\triangle FGH$, hence $\triangle FGH$ is right.
I hope this helps $\ddot\smile$
