Very easy problem : $\angle AEC=\angle C=\angle EFB , \angle BFD=\angle BCD= \angle A - 180+ \angle 2B=\angle B-\angle C=\angle BCF=\angle BDF$ so $BF=BG=BE , BF=BD$ thus $B$ be the circumcenter of $FEDG$ so done.

Like above, a very easy problem, some ideas my repeat from the above solution, since it's not that hard. Extend $DA$.$DA$ and $c_1$ intersect at $T$. 1. $AC=AE$, so $\angle ACE=\angle AEC=\angle EFB$, since $FBCA$ is cyclic, so $EB=EF$ 2. $\angle DEB=\angle DTC=2*\angle DAC=2*\angle DBC$, so $EB=BD$ 3.$EB=BD=BF=BG$...

I didn't prove there is a circumcenter I proved angel FEG=angleFDG