Solution.

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This was a really nice problem..I did it by Ceva in the contest..I was came up with a projective sol..but I wrote the one with with ceva

$\angle FAC=\angle DAC=\angle ACF$ and $\angle CAG=\angle CAB=\angle ACG$. Hence $AFCG$ is a kite and $GF$ bisects $AC$. By symmetry it also bisects $EH$.

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It's easy to see that $\triangle CHG \sim \triangle AEG$. $CA$ is a cevian in both triangles, and it splits $\angle C$ in $\triangle CHG$ up into the same two angles that it splits $\angle A$ into in $\triangle AEG$; thus, it is a corresponding part of the two similar triangles and we have $\frac{HA}{AG} = \frac{EC}{CG}$. By Ceva's Theorem, it is then easy to see that $GF$ bisects $HE$.

Let $L=HE \cap FG$ Let $\angle EAC=\angle CAD=\alpha ; \angle BAF=\angle DAG=\beta$ $\implies \angle FAC=\angle FCA=\alpha; \angle ACD=\angle ACG=\alpha+\beta=\angle GAC$ $\implies GA=GC$ As well: $HA=EC$ By Ceva: $HL=LE$$\blacksquare$

Notice $\triangle AGC$ and $\triangle AFC$ is isosceles with $\angle GCA=\angle GAC$ and $\angle FAC=\angle FCA$. Now notice $\triangle ACH\cong\triangle ACE$ because of one equal side and $\angle FAC=\angle FCA$. This implies $ACEH$ is an isosceles trapezoid. And we are done applying ceva on $\triangle GHE$ withe the concurrency point $F$.

Note that $\triangle AGC$ is an isosceles triangle, then denote $Q = HE \cap FG$ so $AG = AC$. Now doing some angle chasing we can prove that $\angle AHC = \angle ACE$, $ \angle CAH = \angle ACE$ and since they share the side $AC$ and we can conclude that $\triangle AHC \cong \triangle CEA$ $$ \implies AH = CE $$ By Intercept theorem: $$ AC \parallel HE$$ And by Ceva's theorem: $$ QH = QE$$