My solution: Let $O \equiv AC \cap BD$. Then we get that $EAOB$ is a parallelogram and so $\angle EAB = \angle ABO$. Then, we get: $\angle EGF = \angle EAF = \angle EAB + \angle BAF = \angle ABO + \angle BEF = \angle ABD + \angle BEC = \angle ACD + \angle ECA = \angle FCD$ which gives the desired result.

Let $EA,EB$ cut $DC$ at $U,V.$ Then $\angle EAB=\angle ABD=\angle ACD=\angle EVU$ $\Longrightarrow$ $ABVU$ is cyclic $\Longrightarrow$ $CD$ is the image of $\odot(EAB)$ under inversion with center $E$ and power $EA \cdot EU=EB \cdot EV$ $\Longrightarrow$ $D,G$ and $C,F$ are pairs of inverse points under this inversion $\Longrightarrow$ $CDGF$ is cyclic.

We have to prove that the Angels EGF and ECD are equal let p be a point on the line EB out of the segment EB near B we know that EGF=FBP so we have to prove FBP=ECD FBP =BEF+BFE we know that EB is parallede AC so BEF =ECA and we know that EBFGA is cyclic so EFB=EAB we know AE is parallede BD so EAB=ABD finally ABCD is cyclic so ABD=ACD and we are done

$\angle$$BDC$ $=$ $\angle$$BAC =$ $\angle$$EBA$ and $\angle$$EDB =$ $\angle$$AED$ $\longrightarrow$ $\angle$$EFG =$ $\angle$$AED +$ $\angle$$EBA =$ $\angle$$BDC +$ $\angle$$EDB =$ $\angle$$EDC$ $Q.E.D.$

Here is another way of doing the angle chasing: Let $t$ be the tangent to the circumcircle of $ABE$ at $E$. If we can prove that $t \parallel CD$, we are done, because then $\angle GFE = \angle (DE, t) = \angle EDC$, proving the assertion. But because $AEBO$ is a parallelogramm, $t$ is parallel to the tangent to the circumcircle of $AOB$ at $O$ ($O= AC \cap BD$ as in the first solution), and this is parallel to $CD$ because of Reim´s theorem in a limit case, which ends the proof.

$ \angle CDG = \angle CDB + \angle GDB = \angle CAB+\angle DEA=\angle EBA+\angle GFA=\angle EFA+\angle GFA=\angle GFE$ $CFDG$ cyclic.