∠AEB = ∠DAB = ∠ABE (using DA // BE) Triangle ABE is isosceles with angle ABE = angle AEB Let angle ABE = angle AEB = x. Claim: TGNE is cyclic Proof: ∠TNE = 180 - ∠FNB = ∠FGB = ∠EGB = ∠EAB Hence, triangle TNE is similar to triangle BAE (AAA), implying that triangle TNE is also isosceles with ∠NET = ∠NTE = x. ∠ETN = x = ∠ABE = ∠FBN = ∠FGN = ∠EGN. TGNE is cyclic. ∠GDS = ∠GEN = ∠GTN = ∠GTS. Thus, DGST is cyclic (proven).

Claim : $T,G,N,E$ are concyclic Proof. $\angle TEG = \angle AEG = \angle ABG = \angle FBG = \angle FNG = \angle TNG$ We proceed to finish ... $\angle TSD = \angle TNB = 180^{\circ} - \angle TNE = 180^{\circ} - \angle TGE = \angle TGD \implies DGST$ is cyclic. And we are done.

I've got another way to solve it Claim 1: $ATBN$ is cyclic. Proof: $\angle TNB=\pi -\angle EGB=\pi -\angle EAB=\angle TAB$. Claim 2: $AF^2=FG\cdot FD$. Proof: $\angle AGE=\angle ABE=\angle DAF$, then $\triangle AGF \sim \triangle DAF$. Claim 3: $AF^2=FS\cdot FT$. Proof: $\angle ATN=\angle ABE=\angle SAF$, then $\triangle ASF \sim \triangle TAF$. Hence $FG\cdot FD=FS\cdot FT$, hence we are done. $\square$ Well there should be multiple ways to solve it, as the figure contains more properties. For example, i found that $AFGT$ is cyclic, $ENGT$ is also cyclic, and $(TSGD)$ and $(ATBN)$ intersects at another point, which is exactly on the segment $BD$.

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Note: the line $AD$ can be arbitrary and the same angle chase approaches work, it does not need to be a tangent.

It's easy to see that $\triangle ABE$ is isosceles at $A$. We have $\angle{TAG} = 180^{\circ} - \angle{GAE} = \angle{GBE} = \angle{TFG}$. Then $A, T, G, F$ lie on a circle. So $\angle{GTS} = \angle{GAB} = \angle{GEB} = \angle{BEA} - \angle{GEA} = \angle{ABE} - \angle{DAG} = \angle{AGE} - \angle{DAG} = \angle{ADG}$ or $T, D, G, S$ lie on a circle

Also Azerbaijan TST p2

EASY PROBLEM $ BGFN cyclic -> <GNF=<GBF $ $ AGBE cyclic -> <GBF=<GBA=<GEA $ RESULT IS $<GNT=<GET -> GNET cyclic -> <GTN=<GEN$ $DA || BE -> <GEN=<GDA$ RESULT IS $<GTN=<GDA -> <GTS=<GDS -> DGST cyclic$ (IF THERE ARE ANY SAME SOLUTION,SORRY)