AT // BC wanted

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30650 posts

#1 h Sep 22, 2020, 11:26 PM • 13 Y

Y by centslordm, donotoven, jhu08, megarnie, HWenslawski, ImSh95, MightyDog, Amir Hossein, Aopamy, Mango247, Rounak_iitr, deplasmanyollari, ItsBesi

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)

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v_Enhance

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#2 h Sep 22, 2020, 11:27 PM • 25 Y

Y by The_Turtle, Math-Shinai, srijonrick, ProblemSolver2048, v4913, advanture, centslordm, math31415926535, Jc426, donotoven, jhu08, HamstPan38825, ASweatyAsianBoie, rayfish, Lamboreghini, megarnie, ImSh95, crazyeyemoody907, EpicBird08, Amir Hossein, Aopamy, ehuseyinyigit, Aryan-23, H_Taken, NicoN9

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.

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pad

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pad

#3 h Sep 22, 2020, 11:29 PM • 4 Y

Y by centslordm, donotoven, jhu08, ImSh95

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.

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mathlogician

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mathlogician

#4 h Sep 22, 2020, 11:51 PM • 4 Y

Y by ProblemSolver2048, donotoven, jhu08, ImSh95

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 $(CEG).$

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VulcanForge

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VulcanForge

#5 h Sep 23, 2020, 12:12 AM • 4 Y

Y by donotoven, jhu08, ImSh95, NicoN9

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

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PCChess

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PCChess

#6 h Sep 23, 2020, 1:05 AM • 3 Y

Y by donotoven, jhu08, ImSh95

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 $AT \parallel BC$ .

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MP8148

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MP8148

#7 h Sep 23, 2020, 1:18 AM • 2 Y

Y by jhu08, ImSh95

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("$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$", E, dir(45)); [/asy]$
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. $\blacksquare$

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lilavati_2005

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lilavati_2005

#9 h Sep 23, 2020, 1:34 AM • 2 Y

Y by jhu08, ImSh95

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

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Aryan-23

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Aryan-23

#10 h Sep 23, 2020, 2:59 AM • 3 Y

Y by Nathanisme, jhu08, ImSh95

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 . $\blacksquare$

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dchenmathcounts

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dchenmathcounts

#11 h Sep 23, 2020, 5:18 AM • 2 Y

Y by jhu08, ImSh95

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

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djmathman

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#12 h Sep 23, 2020, 5:38 AM • 4 Y

Y by Jc426, jhu08, ImSh95, Mango247

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.

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itslumi

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#13 h Sep 23, 2020, 5:41 AM • 3 Y

Y by jhu08, ImSh95, Mango247

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 $T=T'$ $\blacksquare$

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A-Thought-Of-God

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A-Thought-Of-God

#14 h Sep 23, 2020, 7:55 AM • 2 Y

Y by jhu08, ImSh95

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 $\blacksquare$

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lahmacun

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lahmacun

#15 h Sep 23, 2020, 9:32 AM • 2 Y

Y by jhu08, ImSh95

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 $T'=T$

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geometry6

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geometry6

#16 h Sep 23, 2020, 1:02 PM • 2 Y

Y by jhu08, ImSh95

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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AngeloChu

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AngeloChu

#94 h Jul 13, 2024, 1:54 PM

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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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Mathandski

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#95 h Aug 24, 2024, 2:14 AM

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Subjective Rating (MOHs) $\text{[math]}$

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L13832

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#96 h Sep 10, 2024, 6:01 PM

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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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Yasamin....

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#98 h Sep 13, 2024, 12:01 PM

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sansgankrsngupta

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#99 h Sep 14, 2024, 6:54 AM

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OG!

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), T = dir(70), F = dir(225), G = dir(315), D = dir(185), E = dir(340), B = extension(A,D,F,G), C = extension(A,E,F,G); draw(unitcircle); draw(circumcircle(B,D,F)^^circumcircle(C,E,G)); draw(F--T--G, blue); draw(B--A--C); draw(D--F^^E--G); draw(A--T^^B--C, red); dot("$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$", E, dir(45)); [/asy]$ (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, $AT \parallel BC$

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ezpotd

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#100 h Sep 25, 2024, 6:04 AM

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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.

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SimplisticFormulas

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SimplisticFormulas

#101 h Dec 9, 2024, 9:29 AM

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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 $AT \parallel BC.$ $\blacksquare$

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smileapple

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smileapple

#102 h Dec 27, 2024, 4:08 AM

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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. $\blacksquare$

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Likeminded2017

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Likeminded2017

#103 h Dec 30, 2024, 4:03 AM

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

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Davud29_09

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Davud29_09

#104 h Dec 31, 2024, 8:46 AM

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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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eg4334

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#105 h Jan 1, 2025, 1:52 AM

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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.

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cherry265

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#106 h Jan 25, 2025, 9:19 AM

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Let the line through $A$ parallel to $BC$ meet $\Gamma$ again at $T’.$ A trivial angle chase shows $T=T’$

This post has been edited 2 times. Last edited by cherry265, Jan 25, 2025, 9:20 AM
Reason: small typo

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LeYohan

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LeYohan

#107 h Feb 10, 2025, 10:31 PM

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Easiest G1 ever, just angle chase.

$\angle TGB = \angle GEC$ and $\angle TFC = \angle BDF \implies \angle FTG = 180 - \angle GEC - \angle BDF$ so we now want to proof that $\angle GDF = \angle FTG$ to show that $T$ lies on $(GDF)$ , but just notice that $\angle ADG = \angle GEC$ , hence proved.

Now that we know that $T$ lies on $(DFG)$ , note that $\angle C = \angle TGE = \angle TAC$ , and we're done. $\square$ .

This post has been edited 1 time. Last edited by LeYohan, Feb 10, 2025, 10:31 PM

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Bluesoul

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Bluesoul

#108 h Feb 11, 2025, 1:00 AM

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Connect $AG$ and we have $\angle{AGF}=\angle{BDF}, \angle{DBF}=\angle{AFD}, \angle{DFB}=\angle{DAG}$ , so $\angle{TFG}=\angle{AGF}$ , similar reason after connecting $AF$ yields $\angle{AFG}=\angle{TGF}, \angle{AFT}=\angle{TGA}$ which implies $T$ lies on $(ADFGE)$ , since $\angle{AFG}=\angle{TGF}$ , the $ATGF$ is a isosceles trapezoid which implies the result.

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Maximilian113

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Maximilian113

#109 h Feb 24, 2025, 7:43 PM

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Let $T'$ be on $\Gamma$ such that $AT' \parallel BC.$ We claim that $T=T'.$

Observe that $\angle T'GE = 180^\circ - \angle EGC - \angle AFC = 180^\circ - \angle EGC - \angle GEC = \angle ACB.$ Therefore $(\triangle CEG)$ is tangent to $GT'.$ A similar argument yields $(\triangle BDF)$ is tangent to $T'F,$ so $T=T'$ so we are done. QED

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