just see the power of point $H$ in $\odot$$ABF$ and $\odot$$ABG$

Let $\omega$ be circle of $ABCD$. Draw tangent line to $\omega$ at $A$ and take any point $T$ on this line. Since $GF=GH$ we get $A(T,G;F,H)=-1$. Projecting it to $\omega \implies (A,G;D,B)=-1$. Similarly $(B,G;A,C)=-1$ and projecting it to line $FH$ through $A$ gives $EG=EH$ as desired.

Complex numbers... Let $(ABCD)$ be unit circle and $a=1 , b=-1$ $$e=\frac{d+2cd-c}{c+d} \wedge \bar{e}=\frac{c+2-d}{c+d}$$$$f=\frac{c+2cd-d}{c+d} \wedge \bar{f}=\frac{d+2-c}{c+d}$$Since $H$ is intersection of $AB$ and $FE$ $$h=\frac{\bar{e}f-e\bar{f}}{(f-e)-(\bar{f}-\bar{e})}=\frac{cd+1}{c+d}$$Given that $G$ is midpoint of $FH$ , Hence $$g=\frac{1}{2}(\frac{c+2cd-d}{c+d}+\frac{cd+1}{c+d})=\frac{c+3cd+1-d}{2(c+d)} \wedge \bar{g}=\frac{d+3+cd-c}{2(c+d)}$$Using of the fact that $G$ lies on $(ABCD)$ $$ABGD \text{ is cyclic} \iff \frac{g-d}{(g-1)(d+1)}=\frac{\bar{g}-\bar{d}}{(\bar{g}-1)(\bar{d}+1)} \iff (g-d)(\bar{g}-1)-d(\bar{g}-\bar{d})(g-1)=0 \iff -3cd^2-5d^2+8cd+8d-5c-3=0(\clubsuit)$$$$ABCG \text{ is cyclic}\iff \frac{c-g}{(c-1)(g+1)}=\frac{\bar{c}-\bar{g}}{(\bar{c}-1)(\bar{g}+1)} \iff (g-c)(\bar{g}+1)-c(\bar{c}-\bar{g})(g+1)=0 \iff -3c^2d+5c^2-8cd+8c-5d+3=0(\diamondsuit)$$ $(\clubsuit)$ and $(\diamondsuit)$ yields: $$(c+d)(5c-5d-3cd+3)=0 \text{ if c+d=0 extensions of AD and BC will not intersect , Hence } 5c-5d-3cd+3=0(\spadesuit)$$ Now Let us finish problem $$E \text{ is midpoint of } GH \iff g+h=2e\iff \frac{c+3cd+1-d}{2(c+d)}=\frac{2(d+2cd-c)}{c+d} \iff c+5cd-d+3=4d+8cd-4c \iff 5c-5d-3cd+3=0$$And it is correct by $(\spadesuit)$.....so we are done

Just use Ratio Lemma and just Trig Bash the problem