Solution using barycentric coordinates We take $ABC$ as reference triangle, then $A=(1,0,0); B=(0,1,0); C=(0,0,1)$. Assume that point $E=(0, e, 1-e)$. It is obvious that $F=(e, 0, 1-e), G=(1-e, e, 0)$. Displacement vector $GF$ is $(1 - 2e, e, e - 1)$. It is obvious that $D=( t: 1: 1)$ for some real number $t$. Since $D$ lies on circumcircle of $ABC$, then it is easy to obtain $t=\frac{-a^2}{2b^2}$. So coordinate of $D$ normalized is: $$(\frac{-a^2}{4b^2-a^2},\frac{2b^2}{4b^2-a^2}, \frac{2b^2}{4b^2-a^2})$$. Displacement vector $DE$ is: $$(\frac{-a^2}{4b^2-a^2},\frac{2b^2}{4b^2-a^2}-e, \frac{2b^2}{4b^2-a^2}+e-1)$$So we need to check perpendicularity criterion: $$a^2(e(\frac{2b^2}{4b^2-a^2}+e-1)+ ( e - 1)(\frac{2b^2}{4b^2-a^2}-e))+b^2((1-2e)(\frac{2b^2}{4b^2-a^2}-e) + e(\frac{-a^2}{4b^2-a^2})+(1 - 2e)(\frac{2b^2}{4b^2-a^2}+e-1)+(e - 1)(\frac{-a^2}{4b^2-a^2}))$$The expression above reduces to ( after 5 minutes of algebra): $$(\frac{2e-1}{4b^2-a^2})(2a^2b^2-2a^2b^2)=0$$So we conclude that $DE$ is perpendicular to$FG$ and we are done.

Mine has some calculations too...Please post some good solutions!! let $Z=DE\cap GF, X=AD\cap GF, Y=GF\cap BC$, $M$ the midpoint of $BC$ since $\angle ABD=90$, it suffices to prove that $G, Z, D, B$ are cyclic. let's prove that $\angle BDE=\angle AGF$ $\Leftrightarrow$ prove $\triangle AGX\sim \triangle BDE$ since $\angle BAM=\angle DBM$, let's prove that $AG: GX=BD: BE$ Now some calculation!! let $AG=b, GB=a, \angle BAM=\theta$ note that $BD: BE=AD\sin\theta: 2BG\sin\theta=AD: 2BG$ It suffices to prove $2BG\cdot AG=AX\cdot AD$ Using Menelaus in $\triangle ABC, FGY$ and $\triangle ABM, XGY$ we obtain $\frac {XM}{AX}=\frac {a^2+b^2}{2ab}$ $\therefore AX=\frac {2ab\cos\theta}{a+b}$ $2BG\cdot AG=2ab=AX\cdot AD$ Desired...

Let $O$ be the circumcenter of $\triangle ABC$ Since $AB=AC,$ $GB=GE$ and $FE=FC$ $\Rightarrow$ $AG=EF=FC$ and $AF=GE=GB$ We have: $\triangle GBE \sim \triangle FEC$ so $\frac{GB}{GE}=\frac{FE}{FC} \Rightarrow GA.GB=FA.FC$ So $F$ and $G$ have the same power wrt $(O)$, so $OF=OG,$ combine with $AO$ is angle bisector of $\angle FAG \Rightarrow O$ is the midpoint of arc $FG$ of $(AFG)$ Now use the homothety center $A$ ratio $\frac{1}{2}$ : $D \rightarrow O$, $E \rightarrow M$ with $M$ is the midpoint of $FG$ It's trivial that $OM \perp FG$ so done Note that in general $(AFG)$ passes through Dumpty point of $\triangle ABC$ when $A$ moves on one arc $BC$ of $(O)$ Circumcircle of triangle AQR passes through a fixed point X

Nice synthetic solution, which @Blastoor found during the contest. We start with some simple observations: $BG = GE = AF$ and $GA = FE = FC$ ( this can be obtained doing simple angle chasing ). Let $GE$ intersect $AD$ at point $X$ and $EF$ intersect $AD$ at point $Y$. Simple angle chasing yields : $EX = EY$ and $FY = AF= GE$. So we have $ FC = EF = EY + YF= EX + GE = GX$. Since $GE$ is parallel to $AC$, we conclude that $GFCX$ is parallelogram. $AD$ is diameter and we easily obtain that $GX$ is perpendicular $DC$. Since $AD$ is perpendicular to $BC$ , w conclude that $X$ is orthocenter of triangle $EDC$. So $CX$ is perpendicular to $DE$, but $CX$ is parallel to $GF$, which implies that $DE$ is perpendicular to $GF$ and we are done. Remark: If $GE$ intersect line $AD$ outside circumcircle of triangle $ABC$, then $D$ is orthocenter of $XEC$, then $DE$ is perpendicular $CX$, but since $CX$ is parallel to $GF$, we obtain desired conclusion.

mehhhhh my solution was different from all the above. Let $\angle BAC=a$, and the circunradius $R$. We know $AD=2R$. Note that the perpendicular is true iff $DG^2+EF^2=DF^2+EG^2$, because of pythagoras formula (this is nice because we don't need to deal with the point of contact of the lines). Then, we just need $DG^2-DF^2=EG^2-EF^2=AF^2-AG^2$. But by cossine law applied on $AGD$ and $AFD$ we get: $DG^2=AD^2+AG^2-2.AD.AG.cos a$ $DF^2=AD^2+AF^2-2.AD.AF.cos a$ and subtracting both equations we get $AG^2-AF^2+2.AD.cos a.(AG-AF) = AF^2-AG^2$. Then we just need to have $-2AD.cos a.(AG-AF)=2(AF^2-AG^2) \Leftrightarrow AD.cos a=AF+AG \Leftrightarrow 2R.cos a=AC/sin \angle B=AC/sin 90-a = AC/cos a$, then we just need to get $AC=AF+AG \Leftrightarrow CF=AG=EF$, which is true by the parallelism. $QED$

Here's my "hybrid" solution (just one synthetic step lol). Throw this figure into complex plane where $\overleftrightarrow{AD}$ is the $\textsc{Re}$ axis with $a=1$. Then $e=f+g-1$ and what we are to prove reduces to $g=f\cdot b^2$. But $$\triangle BGE\stackrel{-}{\sim} \triangle CFE$$Or $$\frac{b^2-g}{b^2-f-g+1}=\frac{b^2(fb^2-1)}{(f+g-1)b^2-1}\hspace{2 mm}\leftrightarrow\hspace{2 mm} g=f\cdot b^2\hspace{4 mm} \blacksquare$$

We use moving points, using barycentrics w.r.t. $\triangle ABC$ as our projective coordinates. Move $E$ on $BC$ with degree $1$. $F$ and $G$ move with degree $1$ as well. Thus the displacement vectors $\overrightarrow{DE}$ and $\overrightarrow{FG}$ both have degree at most $1$. By EFFT, the condition $DE\perp FG$ has degree at most $1+1=2$. It suffices to check the problem for three choices of $E$. Here, it is not hard to see that $E=B$, $C$, and the midpoint of $BC$ all work, so we're done.

My solution at https://stanfulger.blogspot.com/2021/08/aops-kite-httpsartofproblemsolvingcomco.html, I hope you will enjoy it! Best regards, sunken rock

Set a=vw ,b=w² and c as v² then d=-vw . Then prove that (f-d)/g-e belongs to real