Geometric problem about incenters 
As shown below, $BC$ is a diameter of $\odot F$. If $AG$ bisects $\angle DAE$ and $FG$ bisects $\angle DFE$, show that $G$ is the I center of $\triangle DEF$.

$\because FD=FE,FG$ bisects $\angle DFE$.
$\therefore GD=GE$.
$\because AD$ bisects $\angle DAE$.
$\therefore ADGE$ con-cyclic or $\triangle ADG\cong\triangle AEG$.
Apparently it’s the former not the latter.
$\therefore\angle GDE=\angle GAE$.
Now we need $\angle FDG=\angle GAE$. I’m not sure whether this contributes to the problem solving.


Please note
$$ \begin {align}
\angle DFE &= 180^\circ - (\angle BFD + \angle CFE) \\
& = 180^\circ - (180^\circ - 2 \angle B + 180^\circ - 2 \angle C)\\
& = 2 (\angle B + \angle C) - 180^\circ\\
& = 2 (180^\circ - \angle A) - 180^\circ\\
& = 180^\circ - 2 \angle A \\
\text {So, }\angle FDE &= \angle DEF = \angle A\\
\end {align}
$$
Alternatively note that $\angle ADE = \angle C$ and so $\angle FDE = 180^\circ - \angle B - \angle C = \angle A$. Similarly, $\angle DEF = \angle A$.
Now given $G$ is intersection of perpendicular bisector of $DE$ and angle bisector of $\angle A$, it must be on the circumcircle of $\triangle ADE$.
That leads to $~\angle GDE = \angle DEG = \angle A/2$
So $DG$ and $GE$ are angle bisector of $\angle FDE$ and $\angle DEF$ respectively.
Therefore we conclude that $G$ is the incenter of $\triangle DEF$.
