If K = AE^GB one can see that B is the orthocenter of triangle AKF. Then KF == d is the polar of P = EG^AB wrt $\Gamma$ and EG is the polar of D wrt $\Gamma$. Therefore DG is tangent to $\Gamma$ at G. It is easy to see that GCDF is cyclic, then <DCF = <DGF. So if G’ = CF^$\Gamma$, it holds that <G’CA = <DCF = <DGF = <GCA, which implies that <G'BA = <GBA or the thesis.

Here is my solution: In the following, we will use directed angles modulo 180° and directed segments (hereby, we will denote directed segments by XY and non-directed segments by |XY|). In order to prove that the reflection H of the point G in the line AB lies on the line CF, we have to prove that < ACH = < ACF. Now, the circle $\Gamma$ is symmetric with respect to the line AB (since AB is a diameter of $\Gamma$). Since the point G lies on $\Gamma$, the reflection H of G in the line AB must therefore also lie on $\Gamma$. Hence, by the chordal angle theorem, < ACH = < ABH. But since the point H is the reflection of G in the line AB, we have < ABH = - < ABG. Thus, < ACH = - < ABG = - < (AB; BG) = < (BG; AB). Since the point G lies on the circle $\Gamma$ with diameter AB, we have < AGB = 90°, or, equivalently, < (AG; BG) = 90°. Hence, < ACH = < (BG; AB) = < (AG; AB) - < (AG; BG) = < (AG; AB) - 90°. On the other hand, < (d; AB) = 90°, since the line d is perpendicular to AB. Thus, < ACH = < (AG; AB) - 90° = < (AG; AB) - < (d; AB) = < (AG; d) = < AFD = - < DFA. Hence, instead of showing < ACH = < ACF, it will be enough to establish - < DFA = < ACF. After the tangent-chordal angle theorem, we have < DEB = < EAB, so that < DEF = < EAB. On the other hand, < EFD = < (BE; d) = < (BE; AE) + < (AE; AB) + < (AB; d). Since < (BE; AE) = 90° (this is because < BEA = 90°, as the point E lies on the circle $\Gamma$ with diameter AB), < (AE; AB) = < EAB and < (AB; d) = 90° (since the line d is perpendicular to AB), this becomes < EFD = 90° + < EAB + 90° = 180° + < EAB = < EAB (directed angles modulo 180° !); together with < DEF = < EAB, this yields < EFD = < DEF. Hence, the triangle DEF is isosceles with |DE| = |DF|. Consequently, $DE^2=DF^2$. But after the intersecting tangent and chord theorem, we have $DE^2=DC\cdot DA$ (with directed segments!). Hence, $DF^2=DC\cdot DA$, and thus DF : DC = DA : DF. Since also, trivially < FDC = - < ADF, we can conclude that the triangles FDC and ADF are oppositely similar. Hence, < DCF = - < DFA, so that - < DFA = < DCF = < ACF, and this completes the proof. Darij

Anyway I found a cute solution using inversion: First of all it’s not difficult to see that: 1. $\angle{EAO}=\angle{OEA}=\angle{EFD}=\angle{DEF}$ hence $ED=DF$. 2. $\triangle{EOD}$ and $\triangle{EAF}$ are similar hence $\angle{EOD}=\angle{DOG}$, and $ED=DG$. From 1. and 2. $DG=DF $. Let $CF$ intersect $\Gamma$ at $G_1$, and $AG_1$ intersect $d$ at $F_1$. Consider the inversion $I$ of pole A and power ${AG}\cdot{AF}$. Since $I(F)=G$ and $I(C)=I(D)$, line $ CF$ transforms to circumcircle of $\triangle{AGD}$ since $F_1$ is on this circumcircle: $\angle{AF_1D}=\angle{DGF}=\angle{DFG}$ hence $AF_1=AF$ and obviously we obtain that $AG_1=AG$ and we are done.

just angle chasing the obvious angles (= and some similar triangles... and done. i suspect its the same as darij's proof above although i haven't read it.

At a QED seminar a nice solution was presented using Pascal's Theorem. Ty it !

do you need some preliminary result? because I couldn't find the hexagon (degenerate or otherwise)... unless I haven't been looking hard enough.

I'm afraid, but I can't find my notes on it right now . As far as I remember you have to apply Pascal at least twice. You're right, that the degenerate version plays an important role; furthermore you also have to consider a point at infinity once. Sorry for the lack of details, but I will try to reconstruct it this afternoon. EDIT: I think I've managed to reconstruct the proof that was given at the seminar (although I can't give a guarantee, that I didn't do a mistake somewhere and that the proof is entirely identical to the original one ) Some Definitions: -$\Phi$ is the intersection of two distinct parallels of $d$ -$Y: =AE\cap BG$ -$S': =AB\cap G'G$ -$F': =EB\cap G'C$ -$G'$ is the intersection of the circle with the parallel of $d$ through $G$. With these definitions it is enough to prove that $G'C$ and $d$ intersect in $F$ First, take the (degenerate) hexagon $AAEBBG$. By applying Pascal we get that $\Phi,Y,F$ are collinear. Thus $Y\in d$ Secondly, take the (degenerate) hexagon $ABGG'EA$. By applying Pascal we get that $\Phi,Y,S'$ are collinear. By the previous we get that $S'$ is on $d$ as well. Thus $S'=S$ Finally, take the (degenerate) hexagon $G'EEBAC$. By applying Pascal we get that $S,D,F'$ are collinear. By the previous we get that $F'$ is on $d$ as well. Thus $F'=F$, which proves the theorem.

We had this at an unofficial tst type test in december. I used 2 inversions to complete my proof then... Anyway, I remember we also had G8 from the isl in our test, but I don't seem to remember it right now and neither find the test sheet, and moreover I don't know if I am allowed to post it.

Let $\angle EAB=\alpha$, $AB$ intersects $d$ at $T$, $G'$ be the refleciton of $G$ by $AB$, $GG'$ intersects $AC$ at $X$. then, cuz $DE$ is tangent to circle $\Gamma$, $\angle DEF= \alpha$ And cuz $\angle AEB= \angle ATF = 90 \DEG$, $A, E, D, F$ are cocircle. so $\angle EFD=\angle EAB=\alpha$. therefore, from $\angle DEF=\angle DFE = \alpha$, we get $DE=DF$ so $DF^2 = DE^2 = DC \times DA$, or ${{DC} \over {DF}}={{DF} \over {DA}} = {{XG} \over {XA}} = {{CX} \over {G'X}}$ cuz ${{DC} \over {DF}} = {{CX} \over {G'X}}$ problem is proved. $Q.E.D.$

There's a cute projective solution. Send $d$ to infinity, in such a way that $\Gamma$ becomes an ellipse, and $AB$ becomes its axis. Then apply the affine transformation that sends $\Gamma$ again to a circle (just scale in the direction of $AB$). Now, we have a circle with two sets of parallel lines, and the result follows trivially (with many little additional corollaries, indeed!).

This problem was used as problem 2 of the final exam of the 3rd TST of Taiwan 2005.

Let O be the center of the circle. Draw OE. $\angle BOE = 2 \angle DEB$. Let AB intersect $d$ at $H$, EHDO is cyclic hence $\angle HDE = \angle HOE =2 \angle DEB$. Now by exterior angle of the triangle DEF we see that $\angle DFE = \angle HDE -\angle DEF = 2\angle DEB- \angle DEB = \angle DEB$ Hence $\triangle DEF$ isosceles so $DE=DF$. By Power of a Point $DE^2 = DF^2 = DC*DA$, or $\frac{DF}{DC}=\frac{DA}{DF}$ Hence SAS similarity $\triangle DFC \sim \triangle DAF$ so $\angle DFC =\angle DAF$. Now since $FC$ intersect $\Gamma$ at another point $X$. $\angle FXG = \angle DAF = \angle DFC$ so $d$ is parallel to $GX$ , which implies $X$ is the reflection of $G$ over $AB$, and the reflection is unique the desired result follows

MindFlyer wrote: There's a cute projective solution. Send $d$ to infinity, in such a way that $\Gamma$ becomes an ellipse, and $AB$ becomes its axis. Then apply the affine transformation that sends $\Gamma$ again to a circle (just scale in the direction of $AB$). Now, we have a circle with two sets of parallel lines, and the result follows trivially (with many little additional corollaries, indeed!). Does anyone know of a decent reference on general projective transformations? I've read some notes on perspectivities (preserving cross ratios) and projectivities, so I've covered the basics of mappings from lines to lines, but I certainly wouldn't be sure that the first transformation given in this post is valid (how do you know there exists a transformation sending d to infinity that keeps a circle as an ellipse?).

Let $ H$ be the intersection of $ FC$ and $ \Gamma$, and let $ I$ be the intersection of $ d$ and $ AB$. By Pascal's theorem on cyclic hexagon $ EEBACH$, we have that $ F$, $ D$, and $ HE \cap AB$ are collinear, that is, we have that $ I$, $ E$, and $ H$ are collinear. Construct point $ F'$ so that $ AIF'$ is directly similar to $ AHB$. Since $ F'I \perp IA$ and $ DI \perp IA$, we have that $ F'$, $ D$, and $ I$ are collinear. A spiral similarity centered at $ A$ maps $ F'I$ to $ BH$, so if we let $ X = F'B \cap IH$, we have that $ ABXH$ and $ AIF'X$ are cyclic. But this means that $ X$ lies on both $ \Gamma$ and $ IH$, implying that $ X = E$. Hence, since $ F'$ lies on both $ BX$ and $ ID$, i.e., both $ BE$ and $ d$, $ F' = F$. Thus, $ \triangle ABH \sim \triangle AFI$. It is clear that $ \triangle AFI \sim \triangle ABG$, so $ \triangle AGB \sim \triangle AHB$. But since the two triangles share side $ AB$, they are congruent; it follows trivially that $ H$, which lies on $ CF$, is the reflection of $ G$ across $ AB$, which completes our proof.

This is triangle geometry in disguise.

that tangent to $\Gamma$ at $E$ intersects midpoint of $HF$. Then $DF^2 = DE^2 = DC\cdot DA$, so $\angle DFC = \angle DAF = \angle GG'C$, so $G'$, $C$, and $F$ are collinear.

Let $G'$ be the image of $G$ with respect to $AB$. Given that $AB$ is a diameter, $G'$ must be in the circle. Consider: $K$ intersection of $d$ and $AB$. $H$ intersection of $AE$ and $d$. We'll build up an inversion relating points. Notice that: $\angle BCA = \angle BCD = \angle DKB = \frac{\pi}{2}$. Hence $KDBC$ is concyclic. Same thing goes for $KBFG$, $KBEH$. Consider the inversion $w$ with center $A$ and ratio $\sqrt{AC*AD}$. Thus $w(E, B, C, G) = H, K, D, F$. Set $G'' = w(G')$. We have $AG' * AG'' = AG*AF => AG'' = AF => AFG'' $isoceles. (1) On the other hand, applying Pascal theorem on $AEECGG$ gives that $DG$ is tangent to the circle. Thus $\angle DGE = \angle A$. Because $EHFG$ is concyclic, we also have $\angle DFG = \angle AEG$. Hence $\angle DGF = \angle DFG$ and so $DGF$ is isoceles. (2) (1) and (2) imply that $DGAG''$ is concyclic. Hence $w(DGAG'') = (G'CF) $ is a line. Or $G', C, F$ are collinear.

I randomly proved that $DG$ is tangent??? Let $G'$ be the intersection of $FC$ and $\Gamma.$ We claim that $DG$ is tagent to $\Gamma.$ Let $G$ be intersection of $AF$ and $\Gamma.$ Note that $\angle EFD=\angle ABF-90^\circ=90^\circ-\angle ABE=\angle BAE=\angle FED$ so $FD=DE.$ Thus, $FD^2=CD\cdot AD$ so $(ACD)$ tangent to $DF.$ Also, $\angle AGC=90^\circ+\angle CAB=\angle ADF$ so $FGCD$ is cyclic. Then, $\angle CGD=\angle CFD=\angle CFD=\angle CAF$ which implies $DG$ tangent to $\Gamma$ as desired. Apply Pascal's theorem on $AGEEBC$ to get $GE\cap BC,D,F$ are collinear, so the intersection of $GE,BC$ lies on $d.$ Then note that Pascal on $BBCG'GE$ gives that the intersection of $GE,BC,$ $F$ and the intersection of $GG'$ and the tangent at $B$ is on $d.$ However, the tangent at $B$ is parallel to $d.$ Thus, $GG'||d,$ as desired.

Let the other tangent from $D$ to $\Gamma$ meet $\Gamma$ at $G'$. Let $AG' \cap BE=F'$ and $AE\cap BG'=X$. Note that by pascal on $AG'G'BEE$ we have that $F', D, X$ are collinear. Furthermore $\angle F'EX= \angle XG'F'=90^{\circ}$ Hence $B$ is the orthocenter of $\triangle AF'X \implies AB \perp F'D\implies F'=F \implies G'=G$. Hence $DG$ is tangent to $\Gamma$. Let $FC \cap \Gamma=H$. Then by Pascal on $CHGGAA$, we have that $F,HG\cap AA, D$ are collinear, but since $FD \parallel AA$, this is possible iff $GH \parallel d \perp AB$ id est $H$ is the reflection of $G $ across $AB$. $\blacksquare$

Let $X$ be the intersection of $AB$ with $d$. Let $P$ be the intersection of $AE$ and $d$. Let $G'$ be the reflection of $G$ around $AB$. Note $\angle AEB=\angle AGB=\angle AG'B=\angle ACB=90$. In addition, because $B$ is the orthocenter of $FPA$, we know $P$, $B$, and $G$ are collinear. We have $\angle APX=90-\angle XAP=\angle EBA$. In addition, because $DE$ is tangent to $\Gamma$, $\angle PED=\angle EBA=\angle APX=\angle EPD$. Thus, $\triangle DPE$ is isosceles. Because $\triangle FPE$ is a right triangle, $D$ is the midpoint of $FP$. Because $\angle BGA$ is right and $P$, $B$, and $G$ are collinear, $\angle FGP$ is right. Since $D$ is the midpoint of $FP$ and $\triangle FPG$ is a right triangle, $\triangle FDG$ is isosceles with $\angle DFG=\angle DGF$. We have $\angle DFG=\angle XFA=90-\angle XAF=\angle GBA=\angle GCA$. So, $\angle GCD=180-\angle GCA=180-\angle DFG$. Therefore, $GCDF$ is cyclic. Therefore, $$\angle DCF=\angle DGF=\angle DFG=\angle GCA=\angle G'CA.$$Therefore, $F$, $C$, and $G'$ are collinear.

Pascal's on $CAGEEB$ to get $\overline{GE} \cap \overline{BC} \in d$. Then, Pascal's on $GG'CBBE$ finishes. Alternatively, let $O$ be the center of $\Gamma$ and $P=\overline{AB} \cap \overline{GE}$. Brokard's on $AGBE$ implies that $F$ and $\overline{AE} \cap \overline{BG}$ form a line perpendicular to $\overline{OP}$, which means $\overline{AE} \cap \overline{BG} \in d$. Then, Pascal's on $ACBGGE$ yields $\overline{AC} \cap \overline{GG} \in d$, i.e. $\overline{DG}$ is tangent to $\Gamma$, hence $AGCE$ harmonic. To finish, let $C \neq Q=\overline{FC} \cap \Gamma$, so we have $$-1=(A,G;C,E)\stackrel{F}{=}(A,G;Q,B),$$which implies $\tfrac{AG}{AQ}=\tfrac{BG}{BQ}$, hence $Q$ is the reflection of $G$ over $\overline{AB}$.

This is fun, to get some motivation its helpful to assume what the problem asks for. Redefine $G$ as the second tangent from $D$ to $\Gamma$, $F$ as the intersection of $AG$ and $BE$, let $BG \cap EA=K$ and $GE \cap BA=J$ and $d \cap AB=X$, now we take polars w.r.t. $\Gamma$ Claim 1: $FK$ is just the line $d$ Proof: By brocard $\mathcal P_J=FK$ and since $J$ lies in $AB$ we have that $\mathcal P_J \perp AB$ and also note that $J \in \mathcal P_D$ so by La'Hire $D \in \mathcal P_J$ and all of those tell us that $FK=\mathcal P_J=d$ Finishing: Now its easy to see that $B$ is ortocenter of $\triangle AKF$ and by ortocenter config we get that $FGEK$ is cyclic and has center $D$ so $DF=DG$, now by the perpendiculars we get $XDCB$ cyclic so by miquel theorem in $\triangle XFA$ we get $FDCG$ cyclic so by arcs we get $\angle DFG=\angle FCD$ so $\angle ACG'=\angle GCA$ which means arcs $AG,AG'$ are equal in $\Gamma$ so $AG=AG'$ and since $AB$ is diameter we get that $G,G'$ are symetric w.r.t. $AB$ thus we are done

Pascal's theorem on $AGEEBC$ shows that $\overline{BC} \cap \overline{GE}$ lies on $d$. [asy][asy] size(10cm); draw(unitcircle); pair O = origin; pair A = dir(180); pair B = dir(0); pair D = (1.6,0.7); pair C = 2*foot(O,A,D)-A; pair M = 1/conj(D); pair E = IP(M--(M+dir(M)*dir(-90)),unitcircle); pair F = extension(B,E,D,conj(D)); pair G = 2*foot(O,A,F)-A; pair G1= conj(G); pair X = extension(C,B,G,E); draw(F--X); draw(A--D--E--F--A--B); draw(G1--F, dotted); draw(G--X--C, lightred+dashed); dot(X); dot("$A$", A, dir(180)); dot("$B$", B, dir(0)); dot("$D$", D, dir(0)); dot("$C$", C, 1.9*dir(85)); dot("$E$", E, dir(-90)*dir(M)); dot("$F$", F, dir(0)); dot("$G$", G, dir(100)); dot("$G'$", G1, dir(-90)); [/asy][/asy] Let $G'$ be the reflection of $G$ over $\overline{AB}$. Then applying Pascal's theorem to $CG'GEBB$ forces $\overline{CG'} \cap \overline{BE}$ to lie on $d$, so the intersection must be the point $F$.

redacted bc fakesolve.

Let $GB \cap AE$ at $P$. Now $AB$ is diameter $FB \perp AP$ and $PB \perp AF$ therefore $B$ is Orthocenter of $\triangle AFP$.As $AB \perp FP$ but $AB \perp \ell$, therefore P lie of line $\ell$. $ED$ is tangent to $(AEBG)$ by Lemma 1.44 (Three Tangents) in EGMO, We get $D$ as midpoint of $FP$ and $DG$ is tangent to circle $(AEBG)$. Its easy to see $FDCG$ is cyclic, $GG' \perp AB$ so we get $\angle DFG = \angle G'GA = \angle GG'A = \angle GCA$ therefore $(FDCG)$ is cyclic. $DF$ tangent to circle we get $\angle CFD = \angle DGC = \angle GAC = \angle GG'C$. $GG'$ is parallel to $FD$ and $\angle DFC =\angle GG'C$ we get $F-C-G$

Denote $G'$ as the second intersection of $FC$ with $\Gamma$. Pascal's on hexagon $EEBACG'$ gives $D$, $F$, and $BA \cap G'E$ collinear, so $BA \cap G'E \in \ell$. Using Pascal's again on hexagon $ABBEG'G$, we have $BA \cap G'E$, $BB \cap GG'$, and $F$ collinear, so $BB \cap G'G \in \ell$. However, since the tangent at $B$ is parallel to $\ell$, it follows that $GG'$ is too, from which we get the desired. $\blacksquare$

Let $R$ be the intersection of $BC$ and $GE$; note that from Pascal on $EEBCAG$, then we find that $R$ lies on $d$. Let $S$ be the intersection of $AE$ and $BG$; note that from Pascal on $AEBBGA$ then $S$ also lies on $d$. Finally, let $T$ be the intersection of $AB$ and $d$. Note that $B$ is the orthocenter of $AFS$, and $T$, $E$, and $G$ are the feet; in particular, $(SF;RT) = -1$. Also note that $D$ is the midpoint of $SF$ from Three Tangents Lemma, a fact I found that we will not use (though it does implies $ACGE$ harmonic). Let $FC$ intersect $(AGCBE)$ at $U$; it suffices to show that $AGBU$ is harmonic. Projecting, \[ (AB;GU) \stackrel{E}{=} (SF;R, UE \cap d) \]So it suffices to show that $U$, $E$ and $T$ are collinear. But this is true from Pascal on $ABCUEG$. $\blacksquare$

Normally I don't post my solutions on AoPS, but I couldn't help but notice the SEVERE lack of complex bashes here. Please enjoy my mathematical equivalent of Star Wars Episode IX.

Some remarks: I saw one earlier complex bash post (#25) by suryadeep. It does mostly the same thing as me (albeit in a different order) except when checking collinearity. I think the key here is getting the most symmetric-looking expression for $f$ possible (particularly, just looking at the expression for $f$ in #25 means that the collinearity check will probably suck, since $f$ is the sum of two fractions). Particularly, noticing that the line doesn't matter anymore is helpful, since the line is mostly just annoying. Finally, the standard way of checking collinearity (determinant) is MUCH more annoying, unless you're OP at row reducing. I think the fact that the denominators of $e + g$ and $e + g + 2$ are similar should inspire just finding the differences of points instead of doing determinant things.

The main idea is that $DG$ is tangent to $(ABC)$. Phantom $G$ to be the other tangent to $(ABC)$ through $D$, and we wish to show that the intersection of $AG$ and $BE$, which we call $F$ lies on $d$. Let $S$ be the intersection of $GE$ and $AB$. Since $S$ lies on the polar of $D$, $D$ also lies on the polar of $S$. This means that the polar of $S$ is $d$, since it is also perpendicular to $AB$ However, by Brocard, $F$ also lies on this polar, thus done. Finally, if we phantom $G'$ is the second intersection of $FC$ with $(ABC)$, then projecting through $F$ gives $$-1=(AC;GE)=(GG';AB),$$so $G'$ is the reflection of $G$ across $AB$ as desired. remark: really, the main idea is that since we are only given one of the two tangents at $D$ to $(ABC)$, we should add in the other one. (This is a good idea in general when you have two objects satisfying very similar properties to add in its "conjugate"). Then, one realizes that the other point is just $G$, and the problem is relatively straightforward from here. remark 2: oh wait, this is really just Pascal twice. oops

Consider a projective transformation taking circle to itself and $k$ to line at infinity. Due to symmethry $AB$ stays the diameter. It is left to prove $G'C \parallel AC$ which is trivial due to symmethry of the new problem.

ok let’s use proj Let $CF$ meet $\Gamma$ again in $H$. Pascals on $EEBACH$ implies that $BA \cap HE \in d$ and Pascals on $EHGABB$ implies that $HG$ and the tangent to $\Gamma$ ast $A$ concur on $d$, so $HG \parallel d$ and we are done.$\blacksquare$