Potla wrote: In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = FG.$ Seems to be wrong. <And confirmed by Geometer's Sketchpad>

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Unfortunately I made a typographical error. The tangent to $\odot(ABF)$ is at $F$, and not $A.$ in other words, $AG$ is tangent to $\odot(ABF)\implies AF=AG;$ and $FG$ is tangent to $\odot(ABF)\implies AF=FG.$ I do not know how this typo merged out. So maybe we have two versions, both of which are true. (1) In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = AG.$ (2) In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $F$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = FG.$ Unfortunately I proved the 2nd version... Can someone verify the first?

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Potla wrote: (1) In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = AG.$ Call the circle on diameter $DE$ as $\omega$. Let the second intersection of $BF,AB$ with $\omega$ be $X,Y$. Now, we have that since $(B,D,C,E)=-1$, therefore, $C$ is the inverse of $B$ wrt $\omega$. Thus, the polar of $B$ wrt $\omega$ passes through $C$. Now, since $C \in AF$ and it is well known that the polar of $B$ passes through $AF \cap XY$, therefore, $X,C,Y$ are collinear. Now, take any point $Z$ on the line $AG$ on the opposite of $A$ than $G$. We have, \[\angle AGY = \angle AFX = \angle BAZ = \angle YAG\] Therefore, $AXGY$ is a cyclic-isosceles trapezium. Now, as $AXFY$ is most obviously an isosceles trapezium, therefore, $\angle CXA = \angle CAX = \angle FYX$. Finally, \[\angle FYA = \angle FYX + \angle XYA = \angle CXA + \angle YXG = \angle AYG\] And hence, $AG = AF$. Potla wrote: (2) In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $F$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = FG$ We employ the same diagram, and take this point to be $G'$. Now, from the result of the previous problem, we have that $angle AXG = \angle AG'F$. Also, we have \[\angle AFG' = \angle ABF = \angle FAG = \angle FXG\] Therefore, we have that \[\angle FAG' = 180^{\circ} -\angle AFG' - \angle AG'F = 180^{\circ} - \angle FXG - \angle AXG = \angle XAF + \angle XFA = \angle XG'A + \angle XG'F = \angle AG'F\] And so, we are done again!

Quick Outline

Rijul saini wrote: Potla wrote: (1) In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = AG.$ Call the circle on diameter $DE$ as $\omega$. Let the second intersection of $BF,AB$ with $\omega$ be $X,Y$. Now, we have that since $(B,D,C,E)=-1$, therefore, $C$ is the inverse of $B$ wrt $\omega$. Thus, the polar of $B$ wrt $\omega$ passes through $C$. Now, since $C \in AF$ and it is well known that the polar of $B$ passes through $AF \cap XY$, therefore, $X,C,Y$ are collinear. Now, take any point $Z$ on the line $AG$ on the opposite of $A$ than $G$. We have, \[\angle AGY = \angle AFX = \angle BAZ = \angle YAG\]Therefore, $AXGY$ is a cyclic-isosceles trapezium. Now, as $AXFY$ is most obviously an isosceles trapezium, therefore, $\angle CXA = \angle CAX = \angle FYX$. Excuse me ,How can you get AXFY is isceles trapezium ???

1) Prove AFYX is an isosceles trapezium A(C,B,D,E)=-1 => A(F,D,Y,E)=-1 => FDYE is harmonic quadrilateral So we have the intersection of two tangents from F,Y lies on ED . But ED is the diameter of (FDYE) So FY is perpendicular to ED . With the same proof we have ADXE is harmonic quadrilateral => AX is perpendicular to ED So FY//AX => AFYX is an isosceles trapezium So angle ABF = 2 GAF . (1) And from AG is the tangent of ABF => ABF= AFG+AGF (2) (1)(2) => AG=AF

Let $\Omega $ denote the $A $-Apollonian circle of $\Delta ABC$. Let $FB\cap \Omega = M $ and $AB\cap \Omega = N $. Since, $F $ lies on $\Omega $, so $FD $ bisects $\angle AFB $. Thus, $A $ and $M $ are symmetrical w.r.t. $DE $. So, $AFNM $ is an isosceles trapezoid. Also, as $AD $ is tangent to $\odot (AFB) $, so $AGMN $ is also a isosceles trapezoid (by easy angle chasing). Hence, $AG = MN = AF $.

Immediatly we see an Apollonian circle.That means that point $A$ belongs to the circle. Denote by $H$ the center of that circle around $\triangle EDA$, we will prove that $H$ belongs to the circle around $\triangle ABF$ We can calculate the angle $\angle DAF = \angle B + \angle C + \frac{1}{2}\angle A = 180 - \frac{1}{2}\angle A \implies \angle DEF = \frac{1}{2}\angle A \implies \angle DHF = \angle BHF = \angle A$, but since $\angle FAB = 180-\angle A $ we get that the quad $FABH$ is cyclic. Up next comes some angle-chase: $\angle DEA = 90 - \angle ADE = 90 - (\frac{1}{2}\angle A + \angle C) \implies \angle AEF = \frac{1}{2}\angle A - \angle DEA = \angle A + \angle C - 90$ Since $G$ belongs to the circle,then we have that $\angle AGF = \angle A + \angle C - 90$ Now that's left to do is calculate $\angle BAG$ We can get with an angle-chase that $\angle HFA = \angle B$ (because of the cyclic quad $FABH$), but we have that $HF=HA$, that means that $\angle FAH = \angle B$ as well. Now we get that $\angle AHF = 180-2\angle B = \angle A + \angle C - \angle B \implies \angle BHA = \angle A - (\angle A + \angle C - \angle B) = \angle B - \angle C $, but since $AG$ is a tangent to the circle around $\triangle ABF$ that implies that $\angle BAG = \angle BHA = \angle B- \angle C$. Now we take a look at the triangle $\triangle GAF$, out of there we see that $\angle GFA = 180 - (\angle BAF + \angle BAG + \angle FGA) = 180-(180-\angle A + \angle B - \angle C + \angle A + \angle C - 90) = 180 - \angle B - 90 = \angle A + \angle C - 90 = \angle FGA$, now that implies that the triangle $\triangle GAF$ is isoceles, which means that $AF = AG$......

Remark

We will use directed angles throughout the solution. Since $F$ is on the $A-Apollonian$ circle, $\angle{ ABE}=\angle {EBF}$. $\frown {AF}=2(\angle{ AFD}+\angle {DAF})=2(\angle {AEF})=\pi-\angle{EBA}$ and $\frown {FG}=\angle{FAG}=\angle{FBA}=2 \angle{EBA}$. Since $\frown{AF}+\frown{FG}+\frown{GA}=2\pi$, $\frown{GA}=\pi-\angle{EBA}=\frown {AF}$.