Let $B_1$ and $C_1$ be the feet from $B$ and $C$ to the opposite sides, respectively. Note that $D, E, F, G$ are invariant under inversion in circle with center $A$ and radius $\sqrt{AC_1\cdot AB}=\sqrt{AB_1\cdot AC}$. Hence $DEFG$ is concyclic with center $A$. Now $CE$ is tangent to $\Omega_1$ since $\angle DEC=\angle DAC=0.5\angle DAE=\angle DGE$. Similarly, $BG$ is tangent to $\Omega_2$. Now the conclusion follows by Pascal on $GGDFEE$.

I think this is quite well known. Claim 1: Points $D,E,F,G$ lie on a circle $\omega$ centered at point $A$. Proof: Let $Y,Z$ be the feet of the altitudes from $B,C$ respectively. Note that $AG^2=AF^2=AZ \cdot AB=AY \cdot AC=AE^2=AD^2,$ and the claim follows $\blacksquare$ Claim 2: $BC$ is the polar of $H$ wrt $\omega$. Proof: Note that $BG,BF$ are the tangents from $B$ to $\omega$ and $H \in GF,$ hence $H$ lies on the polar of $B$ wrt $\omega$. By La Hire's theorem, $B$ lies on the polar of $H$ wrt $\omega$, and similarly $C$ does too, and so $BC$ is actually the polar, as desired $\blacksquare$ To the problem, let $EG,FD$ intersect at $X$. Then, by Brocard's theorem, $X$ lies on the polar of $H$ wrt $\omega$, which is $BC$, as desired.

Let $EG$, $DF$ meet at $I$ and $EF$, $DG$ meet at $J$. By p.o.p, we have $E,F,D,G$ lie on the same circle, denoted by $\omega$. Note that $BF,BG$ and $CD,CE$ are tangent to $\omega$. Applying Pascal’s theorem twice on the hexagons $GGEFFD$ and $DDFEEG$, we obtain that $B,C$ lie on line $IJ$, so we are done.

I will use trinhquockhanh's diagram and point definitions. First of all, $(GDFE)$ is cyclic as $DH\times HE=AH\times HQ=BH\times HX=HF\times HG$ by repeated Power of a Point. Since $AG=AF$, and $AD=AE$, $A$ lies on the perpendicular bisector of $DE$ and $FG$, hence $A$ is the center of that circle. We claim that $AH\times AQ=AG^2$. This is because $AH\times AQ=AY\times AB=AG^2$ by a well known relation of altitudes in right triangles. This means that $Q$ lies on the polar of $H$ wrt to $(GHFE)$, but since we know that the polar of $H$ wrt to $A$ is perpendicular to the diagonal of complete quad $GHFE$ by Brokard, it is precisely $BC$, so we are done. (Note this also proves $GD\cap EF$ lies on $BC$).

Neat problem. We let $H_A$ , $H_B$ and $H_C$ denote the feet of the altitudes from $A$, $B$ and $C$ respectively to the opposite sides. We start off with an easy observation. Claim : Points $D$ , $E$ , $F$ and $G$ lie on a circle with center $A$. Proof : First note that since $AB$ is the diameter of $\Omega_2$, and $GF \perp AB$, it follows that $\triangle AFG$ is isosceles with $AF=AG$. We similarly have that $AD=AE$. Now, note that \[HG \cdot HF = HA \cdot HH_A = HD \cdot HE\]which implies that $DEFG$ is a cyclic quadrilateral. Due to the previous observation, it follows that $A$ must be it's circumcenter. Now, let $X= \overline{GE} \cap \overline{BC}$. We can then notice the following key claim. Claim : Quadrilaterals $XGDH_A$ and $XH_AFE$ are cyclic. Proof : To see why, note that \[2\measuredangle CAD = 2\measuredangle H_BDA = 2(90 + \measuredangle H_BDA) = 2\measuredangle H_BDA = 2\measuredangle EDA = \measuredangle EAD = 2 \measuredangle EGD \]Thus, \[\measuredangle EGD = \measuredangle CAD = \measuredangle CH_AD\]from which it clearly follows that $XGDH_A$ is cyclic. The proof that $XH_AFE$ is cyclic is entirely similar. Now, we are almost there. To finish, note that \[2\measuredangle DXH_A = 2 \measuredangle DGH_A = 2(\measuredangle FGH_A + \measuredangle DGF) = 2\measuredangle FAH_A + \measuredangle DAF\]and \[2\measuredangle FXH_A = 2\measuredangle FEH_A = 2(\measuredangle FED + \measuredangle DEH_A) = \measuredangle FAD + 2\measuredangle DAH_A = \measuredangle FAH_A + \measuredangle DAH_A\]from which we have that $2\measuredangle DXH_A = 2\measuredangle FXH_A$ which implies that $\measuredangle DXH_A = \measuredangle FXH_A$ from which it is clear that $X-D-F$, which proves that the lines $EG, DF$ and $BC$ are concurrent as desired.