Note: B, F, G and C are concyclical. We know that $\frac{CE}{BE}=\frac{DE}{AE}$, as $AF=BE$ and $DG=CE$, $\frac{CE}{BE}=\frac{CE-EG}{BE-EF}\rightarrow CE(BE-EF)=BE(CE-EG)$ $CE(BE)-CE(EF)=BE(CE)-BE(EG)\rightarrow CE(EF)=BE(EG)$, then B, F, G and C are concyclical.

First Solution: We have$$\frac{EA}{AF}=\frac{EA}{EB}=\frac{ED}{EC}=\frac{ED}{EG},$$so $FG\parallel AD$, and therefore $BCGF$ is cyclic. Second Solution: We have\begin{align*}EC\cdot EF & =EC\cdot (EA-AF) \\ & =EC\cdot EA-EC\cdot AF \\ & =EC\cdot EA-EC\cdot EB \\ & =EB\cdot ED-EB\cdot EC \\ & =EB\cdot ED-EB\cdot DG \\ & =EB\cdot (ED-DG) \\ & =EB\cdot EG, \end{align*}so $BCGF$ is cyclic.

FAA2533 wrote: In a cyclic quadrilateral $ABCD$, the diagonals intersect at $E$. $F$ and $G$ are on chord $AC$ and chord $BD$ respectively such that $AF = BE$ and $DG = CE$. Prove that, $A, G, F, D$ lie on the same circle. Bro why are people solving for $BCGF$ being cyclic. When the question asks $AGFD$ to be cyclic

thats cause thats what the actual question is . the aops question is wrong.

ricarlos wrote: Note: B, F, G and C are concyclical. We know that $\frac{CE}{BE}=\frac{DE}{AE}$, as $AF=BE$ and $DG=CE$, $\frac{CE}{BE}=size=200]frac{CE-EG}{BE-EF}[/size]\rightarrow CE(BE-EF)=BE(CE-EG)$ $CE(BE)-CE(EF)=BE(CE)-BE(EG)\rightarrow CE(EF)=BE(EG)$, then B, F, G and C are concyclical.