Let the incircle of $ACD$ touch $CD$ at $H$. Because $HD = EC$, we see that $F$ is the excenter of triangle $ACD$ opposite of $A$. Thus, we have \[ \angle GAF = \angle GCF = \dfrac{\pi}{2} - \dfrac{1}{2}\angle ACD = \pi - \angle ACF = \angle AGF. \]

Yeah I did something on those lines. Ashwath

Still there are no solutions for this problem. I think everybody is waiting for the reply of Darij Grinberg, yetti, Virgil Nicula. Davron Latipov

ThAzN1 wrote: Let the incircle of $ACD$ touch $CD$ at $H$. Because $HD = EC$, we see that $F$ is the excenter of triangle $ACD$ opposite of $A$. Thus, we have \[ \angle GAF = \angle GCF = \frac{\pi}{2} - \frac{1}{2}\angle ACD = \pi - \angle ACF = \angle AGF. \] Davron wrote: Still there are no solutions for this problem. Thazn1 solved the problem, you know! It's very easy...

Please can anyone put a figure of this problem. I can't understand where the point E is located. Davron Latipov

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http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1999Problem6 Vo Duc Dien

$R$ is a point on $AF$ such that $RCD=RAD$ so $ADRC$ is a cyclic quadrilateral. So we have $RC=RD$ also we know $DRC=180-DAC$ and $DFC=90-DAC/2$ Thus $R$ is the center of circumscribed circle of triangle $DFC$ so we have$DR=RC=RF$ Also $FCR=DAG$, $DGA=CFR$ thus $DAG=DGA$ so $GAF=AGF$ QED

See that $F$ is the center of the $A-$ excircle of $\Delta ACD$, consequently $\angle FCD=90^\circ-\frac{m(\widehat{ACD})}{2}$, while $\angle ACF=90^\circ+\frac{m(\widehat{ACD})}{2}$, which will solve the problem. Best regards, sunken rock

By basic properties of incircles we have $CE=\frac{BC+DC-BD}{2}=\frac{AD+DC-AC}{2} \Rightarrow E$ is the point of tangency of the excircle of $\triangle{ADC}$ with $DC$.Now the rest is easy:$\angle{GAF}=\angle{GCF}=180-\angle{ACF}=\angle{AGF} \Rightarrow AF=GF$.

Let the incircle of $\triangle ADC$ touch $DC$ at $H$, and let $F'$ be the $A$-excenter of $\triangle ADC$. Since $\triangle ADC=\triangle BCD$, $E$ and $H$ are reflections over the midpoint $M$ of $\overline{DC}$, hence $E$ is the point of contact of the $A$-excircle of $\triangle ADC$ with $DC\implies EF'\perp CD$. Since both $F$ and $F'$ lie on $AX$, we have $F'\equiv F$. Now letting $X$ be any point on the extension of $BC$ past $C$, we have \[\angle AGF=\angle FCX=\angle FCG=\angle FAG,\] so $\triangle AFG$ is isosceles as desired.

Let $I_1$ the incenter of $\triangle BCD$, $I_2$ the incenter of $\triangle ACD$. So $CDI_1I_2$ is cyclic with center $Q=AI_2\cap (ABC)$. Now perpendicular bisector of $AB,CD$ is also perpendicular bisector of $I_1I_2$ and passes through $Q$, so considering the foot of the altitude from $Q$ to $EF$ we get $QF=QI\implies F$ is the $A$-excenter of $\triangle ACD$ (since it also lies on the angle bisector.) Now it's simple; we want $\angle GAF=\angle FGA\iff \angle GCF=\pi -\angle ACF\iff \angle GCF +\angle ACF=\pi$ which is true since $FC$ is an external angle bisector.

Observe that F is the center of the excircle of $ADC $opposite to $A $(since the center satisfies the two defining properties of $F$). Let line $AC $touch this $excircle $ at $ X$. Then, using fact that $GACF $is cyclic, we have <GAF=<GCF=<XCF=<AGF........hence proved

Is this a joke? Note $E$ is contact point of $A$-excircle of $\triangle ACD$, so $F$ is $A$-excenter, hence $CF$ is external angle bisector of $\angle ACG$ $\implies$ $FA = FG$.

Hmm ... how was this a number 6. Notice that $E$ is the reflection of the point of tangency of the incircle of triangle $ACD$ across the midpoint of $CD$. It is well known that $E$ is the tangency point of the $A$-excircle. Since $F$ lies on the angle bisector of $\angle CAD$ and $FE\perp CD$ it follows that $F$ is the $A$-excenter. So $\angle GAF = \angle GCF = 90 - \frac{1}{2}\angle C$. Also, $\angle AGF = 180 - \angle ACF = 90 - \frac{1}{2}\angle C$. So $\triangle AFG$ is indeed isosceles. $\boxed{}$

My solution. let $L$ be the intersection of the incircle of the triangle $ADC$ with $DC.$ it is easy to see that the triangles $ACD$ and $BDC$ are congruent. hence, we have, $DL=CE.$ and since, $F$ lies on the internal bisector of $\angle DAC$ and $EF \perp CD$ we deduce that, $F$ is the excenter of triangle $ACD$ opposite of $A.$ therefore, \[ \angle FGA = 180 ^\circ -\angle ACF=\angle GCF =\angle GAF\]since, $CF$ is the external bisector of $\angle ACD$ and the quadrilateral $ACFG$ is Cyclic. This implies that triangle $AFG$ is $F$-isoceles.

here is my solution(as beginner) by the properties of incircle and excircle, CE = (BC + CD - BD)/2 = (AD + CD - AC)/2 so, E is the tangency point of the A-excircle of triangle ACD. EF is prependicular to CD deduces F is the excentre of triangle ACD. thus, <GAF = <GCF = 90 - (1/2)*<ACD = 90 - (1/2)*<AFG 2*<GAF = 180 - <AFG; but, <GAF + <AGF = 180 - <AFG so, <GAF = <AGF; AF = GF

it was also given at the book of "Warm up problems for BDMO"

I can't but draw a diagram.

Huh. Maybe I tried too hard on this problem.

Funny problem. Note $ABCD$ is a cyclic quadrilateral. Compute \[\measuredangle AFG=\measuredangle ACG=\measuredangle ACD=\measuredangle BDC\]and \[\measuredangle GAF=\measuredangle GCF=\measuredangle DCF.\]Since $\measuredangle DIF=\measuredangle DIE$ and $\angle DIE=90^\circ-\angle BDC/2$, it suffices to show that $DCIF$ are concyclic. Let $I_B$ denote the $B$-excenter of $\triangle BCD$, then it suffices to show $\angle I_BFI=90^\circ$ by the incenter-excenter lemma. Let $M$ denote the arc-midpoint of arc $CD$ on $(ABCD)$ not containing $A,B$. Observe that the involution of reflection over the perpendicular bisector of $AB$ must take $BM\mapsto AM$ and map $IF$ to the perpendicular line to $AB$ passing through the $B$-Nagel point of $\triangle BCD$, so it maps $F$ to $I_B$ and we are done.

Solution. Just note that $F$ is the $A$-excenter of $\triangle ACD\implies CF$ is an external bisector of $\angle ACG$, let $X$ be a point on line $AC$ beyond $C$, $$\angle FAG=\angle FCG=\angle FCX=\angle FGA.\blacksquare$$

Claim. $F$ is the $A$-excenter of $\triangle ACD$. Proof. This is obvious as $F$ lies on the angle bisector of $\angle CAD$ and $F$ also lies on the perpendicular from $E$ to $CD$, where $E$ is the reflection of incircle touch point with $CD$ over the midpoint of $CD$. $\square$ Now, $$\measuredangle GFA=\measuredangle DCA=2\measuredangle EFC=2(90^\circ+\measuredangle ECF)=2\measuredangle GAF,$$hence $\measuredangle GAF=\measuredangle FGA$, therefore $FG=FA$, we are done. [asy][asy]import geometry; size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,D,e,E,F,G,I; O=(0,0);C=dir(70);D=dir(110); A=dir(195);B=dir(345);path w=circumcircle(A,B,C);e=foot(incenter(A,D,C),D,C);E=2midpoint(D--C)-e; F=intersectionpoint(line(A,incenter(A,D,C)),perpendicular(E,line(E,C))); G=intersectionpoints(C--100D-99C,circumcircle(A,C,F))[1];I=incenter(B,D,C); draw(A--B--C--D--cycle,deep);draw(w,heavyblue);draw(circumcircle(A,C,F),deep); draw(G--F--A,olive);draw(A--G,deep);draw(F--C,deep);draw(F--I,deep);draw(G--D,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$I$",I,dir(I)); [/asy][/asy]

Note that $F$ is the $A$-excenter of $\triangle ACD$. Then, we have \[\angle GAF = \angle GCF = 90 - \frac12 \angle ACD = 180 - \angle ACF = \angle AGF\]Thus, $FA =FG$ and we're done.

Clearly $F$ is the $A$-excenter of $\triangle ACD$. Letting $I$ be the incenter of $\triangle ACD$, we have $\angle ICF=90^\circ$, so $$\angle GAF=\angle GCF=90^\circ-\frac{\angle ACD}{2}=180^\circ-\left(90^\circ+\frac{\angle ACD}{2}\right)=180^\circ-\angle ACF=\angle AGF,$$hence $\triangle AFG$ is isosceles with $FA=FG$. $\blacksquare$

It seems like I’ve found a new solution. (Bc I’m too stupid notice that $F$ is $A$-excenter lol.) Let $I$ be the incenter of triangle $BCD$. Claim: The angle bisectors of $\angle DAC$ and $\angle DBC$ meet on $(ABCD)$. Proof. We use phantom point. Let the angle bisector of $\angle DBC$ intersects $(ABCD)$ at $X$. Note that $X$ is the midpoint of arc $CD$. Hence, $AX$ also bisects $\angle DAC$. Claim: $D,I,C,F$ are concylic. Proof. By Incenter-Excenter Lemma, we have that $X$ is the center of $(DIC)$ so it’s suffice to prove that $XI=XF$. Let $BI$ intersects $CD$ at $Y$. Since the perpendicular bisector of $CD$ intersects $(ABCD)$ at $X$, the perpendicular bisector of $AB$, which is the same as that of $CD$, intersects $(ABCD)$ at $X$ as well, so $XA=XB$. Now, we can do the following angle chasing. \begin{align*} \angle AXB &= 180^{\circ}-2\angle ABY \\ &= 180^{\circ}-2\angle IYE \\ &=180^{\circ}-2(90^{\circ}-\angle YIE) \\ &= 2\angle XIF \end{align*}The claim is now proved. Finishing the problem by angle chasing \begin{align*} \angle{GAF} &= \angle GCF \\ &= \angle DIF \\ &= \angle 90^{\circ}-\angle IDE \\ &=\angle 90^{\circ}-\frac{1}{2}\angle BDC \\ &=\angle 90^{\circ}-\frac{1}{2}\angle ACD \\ &=\angle 90^{\circ}-\frac{1}{2}\angle AFG \end{align*}We are now done.

Note that $E$ is isotomic to the $A$-incircle touchpoint of $\triangle ACD$, which implies that $F$ is the $A$-excenter of triangle $ACD$. So now $$\measuredangle GAF = \measuredangle GCF = \measuredangle FCA = \measuredangle FGA,$$so $FG=AF$.

Note that this configuration implies $F$ is the $A$ ex-center of $\triangle ACD$ Then $$\measuredangle FAG= \measuredangle FCG= \measuredangle ACF= \measuredangle AGF$$Which means $GF=AF$ thus $\triangle AGF$ is isosceles.

Note $F$ is the $A$-excenter of $\triangle ACD$ so \[\measuredangle FGA=\measuredangle FCA=\measuredangle(\overline{CF},\overline{AC})=\measuredangle GCF=\measuredangle GAF\]as desired. $\square$

Sketch: As $ABCD$ is isosceles trapezium $\implies E$ becomes extouch point of $\triangle ACD \implies F$ becomes $A$-excenter of $\triangle ACD$. Now we need to show for $\triangle ABC$ with $A$-excenter $I_A$ that if $P=\odot(ABI_A)\cap BC$, then $I_AA=I_AP$, which we can get by chasing a few more beautiful angels.

Note that $B$ is the reflection of $A$ across the perpendicular bisector of $CD$, therefore $E$ is the point of tangency between the excircle of $\triangle ADC$. Since $AF$ is the angle bisector of $\angle CAD$, then $F$ is the center of the excircle. Define $I$ as the incenter of $\triangle ADC$. Now, simple angle chase will suffice: \begin{align*} \angle AGF = \angle AGC + \angle CGF &= \angle AFC + \angle CAF \\ &= 180 - \angle ACF = 180 - \left(ICF + \angle ACI\right) = 90 - \tfrac{\angle ACD}{2}. \end{align*}In addition, $\angle AFG = \angle ACG = \angle ACD$. Thus, \begin{align*} \angle GAF = 180 - \angle AGF - \angle AFG = 180 - \left(90 - \tfrac{\angle ACD}{2}\right) - \angle ACD = 90 - \tfrac{\angle ACD}{2} = \angle AGF. \end{align*}Thus, $\triangle AFG$ is isosceles, as desired. OoPsOoPs.

Notice that since $ABCD$ is an isosceles trapezoid, the reflection of $E$ across the midpoint of $CD$ lies on the incircle of $\triangle ACD$ by symmetry. So it follows that $E$ is the ex-touch point. And since $F$ lies on the angle bisector of $\angle DAC$ and $E$ is the projection of $F$ onto $CD$, $F$ is the $A$-excenter wrt $\triangle ACD$. Then $\angle GAF = \angle GCF = 90^{\circ} - \frac{\angle DCA}{2}$ and $\angle FGA = 180^{\circ} - \angle FCA = 90^{\circ} - \frac{\angle DCA}{2}$ so $\angle FGA = \angle GAF \implies \triangle FGA$ is isosceles as desired.

$\angle AGF = \pi - \angle ACF = \angle DCF = \angle GCF = \angle GAF$. $\square$

$F$ is the excenter of $\triangle ACD$ and thus $F$ is the external bisector of $\angle ACG$ and we are done as $\angle GAF=\angle GCF=180-\angle ACF=\angle FGA.$

Reflecting $E$ over the perpendicular bisector of $CD$ yields the touch-point of the incircle of $\triangle ACD$ and $CD.$ Therefore $F$ is the $A$-excentre of $\triangle ADC$ so $$\angle GAF = \angle DCF = 180^\circ - \angle FCA = \angle AGF.$$QED