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.

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.
