if $G_{1}$ is the midpoint of $DE$ prove that $AG_{1}$ is perp to $BE$ using the midpoint of $BD$ and some similar triangles.

My solution : Consider the circles : $C_1$ by diameter $[AB]$ and $C_2$ by diameter $[AE]$. We have $\angle{ADE}\equiv \angle{ACD}\equiv\angle{ABC}$, so $DE$ is tangent at $C_1$. Obviously, $DE$ is tangent at $C_2$, because $DE\perp AE$. So $DE$ is common tangent from the circles $C_1$ and $C_2$. But $AF$ is radical axis from this circles. By the power of point, the intersection $AF\cap DE$ is the midpoint of $[DE]$.

This solution is not quite different, but still... Obviously since $F=BE\cap\odot(ABD)$ therefore $AF\perp FE.$ So $GE$ is tangent to $\odot(AFE),$ and hence $GE^2=GF\cdot GA.$ Again, if $M$ is the midpoint of $AB$ then $MD\parallel AC$ and $DE\perp AC$ implies that $MD\perp DE.$ Since $M$ is the centre of $\odot(ABD),$ therefore $ED$ is also tangent to $\odot(ABD).$ Hence $GD^2=GF\cdot GA.$ So, $GE^2=GD^2$ and hence $G$ is the midpoint of $DE.\Box$

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The problem can be generalized: It's just consequence of the third post here. Best regards, sunken rock

let $G$ be $O$ now suppose $O$ is midpoint of $DE$ we prove that $AO$ and $BE$ are perpendicular. set $E$ as origin $(0,0)$ and let $DE$ be $Y $axis and $AC$ be the $X$ axis. let coordinates of $D=(0,U), A=(V,0), C=(-W,0)$ than as $O $is mid point of $DE$ thus coordinates of$ O=(0,U/2)$ AND COORDINATES OF $B=(W,2U)$ as $D$ is midpoint of $BC.$ now $AB = AC$ gives us $VW = U^2 $ thus product of slopes of$ AO$ and $BE$ is$ = (-U/2V)(2U/W) = -U^2/VW = -1$ since we have$ VW = U^2$ hence $AO$ and $BE$ are perpendicular.

Make $M$ the midpoint of $AB$, then $M$ is the circumcenter of triangle $ABD \implies MD$ is medium base of triangle $ABC \implies$ $MD||AC$, as $DE$ is perpendicular to $AC \implies DE$ is perpendicular to $MD$, so $DE$ is tangent to the circumcircle of $ABD$ in $D$. See that quadrilateral $ABDF$ is cyclic, thus, $AFB = ADB = 90º \implies AFE = 90º$, and from this we get that the circumcenter of triangle $AFE$ is at $AC$, as $AC$ is perpendicular to $DE$ at $E \implies$ the circumcircle of $AFE$ is tangent to $DE$ at $E$. So $AF$ is the radical axis of the circumcircles of $ABD$ and $AFE \implies G \in AF \implies$ we saw that $GD$ and $GE$ are tangent to the circumcircles of $ABD$ and $AFE$ respectively $\implies$ $GD^2 = GE^2 \implies GD = GE. \blacksquare$

For variety, here's another way via similar triangles: Let $H$ be a point on $AD$ such that $HG\perp DE$. $\angle{DHG}=\angle{CDE}$ and since $AFDB$ is cyclic, $\angle{HAG}=\angle{DBE}$, so triangles $CDE$ and $DHG$ are similar and triangles $HAG$ and $DBE$ are similar. So, since $D$ is the midpoint of $BC$, $H$ is the midpoint of $AD$. Since $HG\parallel AE$, by Midpoint Theorem, $G$ is the midpoint of $DE$.

Let $\angle{DAG}=\angle{X}$ and $\angle{GAE}=\angle{Y}$ . Then $\angle{DAB}=\angle{X+Y}$ . Since $\angle{ADB}=\angle{AFB}$ and $D$ is midpoint of $BC$ , $\angle{AFB}=\angle{90}$ . $\angle{BAF}=\angle{FDE} + \angle{EDC} = \angle{FDE} + \angle{X+Y}$ , $\angle{FDE}= \angle{X}$ . So we get that $DG^2=FG*GA$. And on the other hand we have that $\angle{FED} = \angle{BFD} - \angle{FDE} = \angle{Y}$ , which means $GE^2=FG*GA$. So $DG=GE$ .