We have $$\measuredangle TGF = \measuredangle TGC = \measuredangle GEC = \measuredangle GEA = \measuredangle GFA$$ and similarly $\measuredangle TFG = \measuredangle FGA$. Thus $TAGF$ is an isosceles trapezoid, as needed.

Diagram

Define $\omega=(ADE)$, and let $T'=FF\cap \omega$. Then $180-\angle T'AD=\angle T'FD=\angle DBF$, so $\angle T'AD=180-B$. This means $AT'\parallel BC$. If we let $\ell$ be the line through $A$ parallel to $BC$, then this means $T'=\ell \cap \omega$, i.e. $FF, \ell, \omega$ concur. By symmetry, $GG,FF,\ell,\omega$ concur, which means $T=FF\cap GG \in \ell$, as desired.

Consider $T'$ such that $AFGT$ is a cyclic isosceles trapezoid. We contest that $T = T',$ so it suffices to prove that $\overline{T'F}$ and $\overline{T'G}$ are tangent to $(BDF)$ and $(CEG),$ respectively. Indeed, check that $\angle DFT' = \angle 180 - \angle DAT' = \angle ABC = \angle DBF,$ so $\overline{T'F}$ is tangent to $(BDF).$ Similarly $\overline{T'G}$ is tangent to $(CEG).$

Redefine $T$ so $AT \parallel BC$ and $T \in \Gamma$; we show this is the desired point. Indeed, we have $$\begin{align*} \measuredangle TGE &= \measuredangle TAE = \measuredangle GCE \\ \measuredangle TFD &= \measuredangle TAD = \measuredangle FBD \\ \end{align*}$$

Similar to post 2, oh well. Let $P$ be a point on $\overline{TF}$ such that $F$ lies between $P$ and $T$ and $Q$ be a point on $TG$ such that $G$ lies between $T$ and $Q$. Observe that $$\measuredangle TGF = \measuredangle TGC=\measuredangle QGC=\measuredangle GEC=\measuredangle GEA=\measuredangle GFA.$$ Further, $$\measuredangle TFG=\measuredangle PFB=\measuredangle FDB=\measuredangle FDA=\measuredangle FGA.$$ From these 2 equations, one gets that $\measuredangle TGA=\measuredangle TFA$, so $ATFG$ is cyclic. Since we've established that $\measuredangle GFA=\measuredangle TGF$, $ATGF$ is an isosceles trapezoid which implies that $AT \parallel BC$.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("$A$", A, dir(110)); dot("$T'$", T, dir(70)); dot("$B$", B, dir(225)); dot("$C$", C, dir(330)); dot("$F$", F, dir(270)); dot("$G$", G, dir(270)); dot("$D$", D, dir(135)); dot("$E$", E, dir(45)); [/asy][/asy] Let $T'$ be the point on $\Gamma$ such that $\overline{AT'} \parallel \overline{BC}$. I claim that $T=T'$. Angle chasing gives $$\angle B = \frac{1}{2}(\widehat{AEG} - \widehat{DF}) = \frac{1}{2}(\widehat{T'DF} - \widehat{DF}) = \angle DFT',$$ so $\overline{T'F}$ is tangent to $(BDF)$ and similar for $\overline{T'G}$, implying the conclusion. $\blacksquare$

Diagram

Redefine $T$ as the point on $(ADFGE)$ such that $AX \parallel BC$. $$\angle AXF = \angle BDF = 180 - \angle XFB \Longrightarrow \text{XF is tangent to} (BDF)$$ $$\angle XAC = \angle XGE = \angle ACB \Longrightarrow \text{XG is tangent to} (CEG)$$

Redefine $T$ to be the point on $\odot(ADE)$ such that $AT\parallel BC$. For brevity write $\angle ATF = \angle AEF =\angle BDF = \theta$. So we have $\angle ABC+\theta + \angle DFB =\pi$. We also have $\angle DFB+\angle DFE+\angle DFG - \pi \implies \angle TFD = \angle ABC$. This means that $TF$ is tangent to $\odot (BDF)$. Similarly $TG$ is tangent to $\odot (GEC)$ and we are done . $\blacksquare$

fun So note that $$\measuredangle GFT=\measuredangle BFT=\measuredangle BDF=\measuredangle ADF,$$ and similarly, $$\measuredangle TGF=\measuredangle GEA.$$ And also note that $\measuredangle GEA+\measuredangle ADF=\measuredangle GAF,$ and $\measuredangle GFT+\measuredangle TGF=\measuredangle GTF.$ So $T$ lies on $(ADFGE)$. Now note $$\measuredangle ATF=\measuredangle ADF=\measuredangle BDF=\measuredangle GFT,$$ which implies parallelism. gg

Note that $\angle DFT = \angle ABG$ by the tangency condition and $\angle DFB = \angle BAG$ by cyclic quadrilaterals, so $\angle TFG = \angle AGF$. Likewise, $\angle TGF = \angle AFG$, so triangles $AFG$ and $TGF$ are congruent and we win.

Let the tangent at F to $(BDF)$ intersect $(ABC)$ at a point $T'$. Now prove that AT' parallel with BC. $\angle DBF=\angle DFX$ Since $(AT'FD)$ cyclic we know that $\angle DAT'=\angle DFT'$, Which means that $\angle T'AC=\angle ACB$ ,which means that $AT'\parallel BC$. Now prove that $T'G$ is tangent to $(GEC)$, or we can prove that $\angle T'GE=\angle ECG$. $\angle T'AE=\angle ACB'$ now since $(AT'EG)$ cyclic, we have $\angle T'AE=\angle T'GE'$, which shows us that $AT'\parallel BC$. and tanget to $(BDF)$ at $F$ and to $(GEC)$ at $G$, Wich shows that $T=T'$ $\blacksquare$

Let $P$ be a point such that $AFGP$ is Isosceles trapezoid then clearly $P \in \odot ADFGE$ and that $AP \parallel BC$. Now $$\begin{align*} \angle DAP &=\angle A + \angle EAP \\ &=\angle A + \angle CAP \\ &=\angle A + \angle ACB \\ &=\angle A + \angle C \\ &\implies \angle DFP=\angle B \\ \end{align*}$$ which shows that $PF$ is tangent to $\odot BDF$ and similarily $PG$ is tangent $\odot CEG$ so $P \equiv T$ which implies the desired result $\blacksquare$

Let the line through $A$ parallel to $BC$ meet $\Gamma$ at $T'$. $$\measuredangle T'FD=\measuredangle T'AD=\measuredangle T'AB=\measuredangle CBA=\measuredangle FBD$$ So, $T'F$ is tangent to $(BDF)$ and similarly $T'G$ is tangent to $(CEG)$ and $T'=T$

My solution posted for storage: Let $T'$ be the point on $\Gamma$ such that $\overline{AT'} \parallel \overline{BC}$. Claim 1. $T'\equiv T$. Proof. We need to prove that tangent from $G$ to $(EGC)$ and tangent from $F$ to $(DBF)$ meet at $T' \iff \angle ACB= \angle EGT'$ and $\angle ABC= \angle DFT'$. Since $\overline{AT'} \parallel \overline{BC} \implies \angle ACB=\angle TAC=\angle TAE=\angle TGE \implies T'G$ is tangent to $(EGC)$ and similarly $T'F$ is tangent to $(DBF) \implies T'\equiv T.\blacksquare$ Now, by the definition of $T'$ and Claim 1. $\implies AT \parallel BC.$$\square$

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The main claim is that $T \in (ADFGE)$. To show this, let $H=AE \cap FT$ and note that Pascal's Theorem on $TFGAEA$ gives $H, C, A$ collinear. This is clearly true, so $T \in (FGAE)$ by the converse of Pascal, as desired. To finish, note that $\angle{DFT} = \angle{ABC}$, so $\angle{TAC}=180-\angle{ABC}-\angle{BAC}=\angle{BCA}$, implying $AT \parallel BC$, as desired.

Sorry for eye touring, It seems I didn't use latex correctly Let H be the intersection of the circle (ADFGE) and the tangent line of circle (BDF) passing through D. Since $\angle DBF= \angle FDA= \angle FHA$, we conclude that $AH \parallel BC$. Now we will show that HG is tangent to circle (GEC), this is because $\angle HGE= \angle HFE=180^{\circ}- \angle FHE- \angle FEH$, where $\angle FHE= \angle FAE= \angle EGC$, and $\angle FEH= \angle FGH= \angle GHA= \angle AFG= \angle GEC$, so we prove that HG is tangent to circle (GEC), so H=T, and we has proved that AH is parallel to BC, as desired. LaTeX fixed. You were adding extra dollar signs to the beginning of the angle commands. (Click on the LaTeX to see how it's done) ~dj

Let $S$ be on $\Gamma$ such that $FS$ is tangent to $(BDF)$. Then by tangency $\angle GBA=\angle DFS$. Also, $\angle DFB=\angle BAG$ by antiparallelity. Hence $\angle SFG=\angle AGB\stackrel{\text{cyclic}}{=}\angle ASF \implies AS \parallel BC$. From the symmetry the tangent to $(CEG)$ at $G$ passes through $S$ and hence $S \equiv T$, as desired. $\blacksquare$

One of the easiest problems I have seen in recent years...it's even easier than our nation's nationals

kvs wrote: The main claim is that $T \in (ADFGE)$. To show this, let $H=AE \cap FT$ and note that Pascal's Theorem on $TFGAEA$ gives $H, C, A$ collinear. This is clearly true, so $T \in (FGAE)$ by the converse of Pascal, as desired. To finish, note that $\angle{DFT} = \angle{ABC}$, so $\angle{TAC}=180-\angle{ABC}-\angle{BAC}=\angle{BCA}$, implying $AT \parallel BC$, as desired. I think your solution is wrong...Because converse of pascal logic dosent seem right...as a counter example.....Let $P$ be a point such that $P$ dosent lie on $\odot AFGE$ ....By Pascals theorem on $PFGAEA$, we get $PF\cap AE,C,A$ are collinear..Then by coverse of Pascals theorem ,$P$ lies on $\odot AFGE$ (ACCORDING TO YOUR LOGIC)...AND ITS DEFINITELY NOT TRUE

Let $FT \cap \Gamma = V$. Then $$\measuredangle VAB = \measuredangle VAD = \measuredangle VFD = \measuredangle FBD = \measuredangle BCA$$ $$\implies AV \parallel BC.$$ Similarly, If $GT \cap \Gamma = U$, then $$AU \parallel BC$$ . Therefore, we get $U = V = T$, and in turn $AT \parallel BC$. $\square$

Let $T'$ be the intersection of $(ADE)$ and the line through $A$ parallel to $BC$. We can say: $$\angle ABC=180-\angle T'AD=\angle DFT',$$ $$\angle ACG=\angle T'AE=\angle T'GE,$$ which implies that $T'F$ tangent to $(BDF)$ and $T'G$ tangent to $(CGF)$, respectively. Therefore, $T=T'$, as desired $\blacksquare$

Let the circle passing through $A$, $D$, $E$, $F$, and $G$ be $\omega$. I claim that $T$ is the unique point on $\omega$ such that $AT\parallel BC$. I will prove this using a phantom point argument. Let $T'$ be the unique point on $\omega$ such that $AT'\parallel BC$. Note that this implies that $$AT'\parallel BC \iff AT'\parallel FG,$$ which means that $AT'GF$ must be a isosceles trapezoid, since $AT'FG$ is cyclic. This implies that $$\angle T'GF=\angle AFG=180-\angle AEG=\angle GEC,$$ which in turn gives us that $$\angle T'GE=180-(\angle EGC+\angle T'GF)=180-(\angle EGC+\angle GEC)=\angle ECG,$$ which by tangent properties, implies that $T'G$ is tangent to $(EGC)$. Similarly, we also have that $$\angle T'FG=\angle AGF=180-\angle ADF=\angle BDF,$$ which then gives that $$\angle T'FD=180-(\angle DFB+\angle T'FG)=180-(\angle DFB+\angle BDF)=\angle DBF,$$ which similarly implies that $T'F$ is tangent to $(BDF)$. Therefore, since $T'F$ is tangent to $(BFD)$ and $T'G$ is tangent to $(CGE)$, this means that $T'=T$. This implies that $AT\parallel BC$, finishing the problem.

parmenides51 wrote: Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$. (Nigeria) Let $T$ be on $(ADE)$ such that $AT//BC$ then we have: $\angle ACB=\angle TAC=\angle TAE=\angle =\angle TGE$ so $TG$ is tangent to $(CEG)$ SIMILAR FOR $(BDF)$

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literal minute solve (i took 2 minutes cuz i dumb) let $U$ be on circle $\Gamma$ that $AT$ and $BC$ are parallel $UAE=UGE$ by cyclicity, and $UAE=GCE$ by parallel then $UGE=ECG$ so $UG$ is tangent to $(GEC)$ similarly, we can prove $UF$ is tangent to $(BDF)$ so we have that $U=T$ thus, $AT$ is parallel to $BC$

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Solution with @Mathematical Arceus Claim I: $\odot(FTAG)\iff \measuredangle GTF = \measuredangle GAF$ Proof: $$\begin{align*}\measuredangle GTF & = \measuredangle GFT + \measuredangle TGF\\&=(\measuredangle TFD + \measuredangle DFB) + (\measuredangle TGE + \measuredangle EGC)\\&= (\measuredangle DBF + \measuredangle BFD)+ (\measuredangle EGC + \measuredangle ECG)\\&= \measuredangle BDF + \measuredangle CEG\\&= \measuredangle FGA + \measuredangle AFG\\ &= \measuredangle GAF \end{align*}$$ Now by applying Reim's Theorem on $\odot (DFB)$ and $\odot(TGFA)$ we get that $AT \parallel BC$.

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OG! [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("$A$", A, dir(110)); dot("$T$", T, dir(70)); dot("$B$", B, dir(225)); dot("$C$", C, dir(330)); dot("$F$", F, dir(270)); dot("$G$", G, dir(270)); dot("$D$", D, dir(135)); dot("$E$", E, dir(45)); [/asy][/asy] (FIGURE COPIED FROM MP8148) Now $\angle ECG=\angle EGT=\angle EAT$( The first equation follows from Tangent-Secant Theorem, the second statement follows from the fact that $A,T,E$ and $G$ are cyclic.) Thus, $AT \parallel BC$

ok fine. FINE. after years of not caring ill finally do it. average nigerian problem. well well well. We claim that $T$ is the reflection of $A$ over the perpendicular bisector of $FG$. It suffices to prove $\angle TFG = \angle AGF, \angle TGF = \angle TFG$. We prove the former and the latter follows by symmetry. $\measuredangle TFG = - \measuredangle TFB = \measuredangle FDB = -\measuredangle ADF = \measuredangle AGF$, so we are done.

Simple stuff. Let the tangent to $\odot (BDF)$ at $F$ meet $\odot (ADE)$ again in $T’$. Indeed, $\angle B=\angle T’FD=\angle T’ED$ and $\angle DT’E=\angle DAE=\angle A$, so $\angle TGE=\angle C=\angle T’DE=\angle T’GE \implies T’ \equiv T$. Finally, $\angle TAC=\angle TAE=\angle TDE=\angle C=\angle ACB$, so $AT \parallel BC.$ $\blacksquare$

Let $U$ be a point on the extension of segment $TF$ such that $F$ lies between $T$ and $U$. Then $\angle TFC=\angle BFU=\angle BDF=\angle ACF$, where the second equality follows due to inscribed angles in the circumcircle of $\triangle BDF$ and the third equality follows from the cyclicity of $ADFC$. Similarly $\angle TCF=\angle AFC$. Thus $ATCF$ is an isosceles trapezoid as desired. $\blacksquare$

$180-\angle ATG=\angle CFA=\angle CEG=180-\angle TGC=\angle TGF$ so $FG \parallel AT$ and thus $BC \parallel AT.$

Tangent to (BDF) at F intersect Γ at T. Claim 1:AT//BC.Proof:AB intersect FT at X.From tangent <BDF=<BFX=<TFG=α.From ATFD is cyclic <ATF=<BDF=α ,and we get <ATF=<TFG=α we proved claim 1. Claim 2:GT tangent to (GEC). Proof: We note that <FTG=β from ATEG is cyclic we get <ATG=<AEG=α+β.Then we get <GEC=180-α-β.From claim1 <TGF=180-α-β.And <(BC,TG)=180-α-β from this claim 2 is proved.We are done.

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$T$ can lie on either side of $A$, but the cases are symmetric so for notation let $T$ be on the side of $C$. Let $O$ be the center of $\Gamma$. We first prove that $T$ is on $\Gamma$. This is because the $F$ tangent intersects $\Gamma$ at $T'$ such that $DAT' = 2 \angle B$. The $G$ tangent intersects it such that $ET'' = 2 \angle C$. But $DAT' + ET'' = 2 \angle B + 2 \angle C = DAE$ thus $T' = T''$. Note without the angle symbol we are talking about arcs on $\Gamma$. From here, $\angle B = \angle DFT = 180 - \angle DAT$ which finishes.