$ABCD$ is cyclic, so lines $FE$ and $FG$ are isogonal wrt angle $AFB$. Thus if $FG\cap EC=Z$, we have \[\angle FHD=\pi - \angle HFD -\angle HDF=\pi -\angle ZFC -\angle ZCF=\pi - \angle DGF.\]

Burii wrote: $ABCD$ is cyclic, so lines $FE$ and $FG$ are isogonal wrt angle $AFB$. Thus if $FG\cap EC=Z$, we have \[\angle FHD=\pi - \angle HFD -\angle HDF=\pi - \angle HFD -\angle ZFC -\angle ZCF=\pi - \angle DGF.\] easy question

Burii wrote: $ABCD$ is cyclic, so lines $FE$ and $FG$ are isogonal wrt angle $AFB$. Thus if $FG\cap EC=Z$, we have \[\angle FHD=\pi - \angle HFD -\angle HDF=\pi - \angle HFD -\angle ZFC -\angle ZCF=\pi - \angle DGF.\] Wait why is this? In your solution, you are suggesting that $\angle HDF=\angle ZFC+\angle ZCF$, but don't we know that $\angle ZCF=\angle BDF=\angle HDF$ by the properties of cyclic quads and reflections?

At MOP, robinpark complex bashed this problem.

He always bashes

Fixed similarity signs.

Let $ AH $ and $ CG $ intersect at $ I $. Then we get $ DHIG $ to be cyclic because $ \angle DHI + \angle DGI = \angle DEC + \angle DEA = \pi $. Note also that $ \angle FCI = \angle DBC = \angle DAC = \angle DAH = \angle FAI $. Hence we get $ ACIF $ cyclic and $ \angle HIF = \angle ACF = \angle ADB = \angle HDF $. Therefore, $ F $ is on the circumcircle of $ DHIG $ and the conclusion follows.

A solution requiring no additional points: We easily find $\triangle FAB \sim \triangle FCD$ and $\triangle EAB\sim \triangle GCD$. From these it follows that $\triangle FBE\sim\triangle FDG$. We now have $\angle FHD=\angle FED=180^\circ-\angle FEB=180^\circ-\angle FGD$, so $FHDG$ is cyclic.

i have another beautiful solution! My solution: Let $EH \cap AD=P(.)$ and $M(.)$ lies on the line $DA$ such that $MP=PD$. So, easily get $FMEC$ cyclic and $MEDH$, $MCGH$ are parallelogram. $ \angle HFD= \angle MFE= \angle MCE= \angle HGD$ $ \Rightarrow $ $H,F,G,D$ are concyclic.

[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.56, xmax = 25.44, ymin = -5.49, ymax = 13.49; /* image dimensions */ pen ffxfqq = rgb(1,0.5,0); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1); /* draw figures */ draw((-3.236895072206104,1.716522681043244)--(2.195238943204361,2.358939626736120)); draw((2.055826808152264,0.02425299063591571)--(-3.236895072206104,1.716522681043244)); draw((-0.4580563678168402,2.045154682386091)--(-0.6131947965902200,0.8776332488183485)); draw((2.195238943204361,2.358939626736120)--(2.055826808152264,0.02425299063591571)); draw((-0.4580563678168402,2.045154682386091)--(2.055826808152264,0.02425299063591571)); draw((-0.6131947965902200,0.8776332488183485)--(2.195238943204361,2.358939626736120)); draw((2.195238943204361,2.358939626736120)--(3.893636125792478,0.9936042076656184)); draw((3.893636125792478,0.9936042076656184)--(2.055826808152264,0.02425299063591571)); draw(circle((0.9800000000000012,1.260000000000001), 1.638436447437424)); /* dots and labels */ dot((2.195238943204361,2.358939626736120),dotstyle); label("$D$", (2.256494727886761,2.450823303759719), NE * labelscalefactor); dot((-0.4580563678168402,2.045154682386091),dotstyle); label("$A$", (-0.3928179596270162,2.144544380347722), NE * labelscalefactor); dot((-0.6131947965902200,0.8776332488183485),dotstyle); label("$B$", (-0.5459574213330147,0.9653705252115314), NE * labelscalefactor); dot((2.055826808152264,0.02425299063591571),dotstyle); label("$C$", (2.118669212351362,0.1231034858285383), SE * labelscalefactor); dot((-3.236895072206104,1.716522681043244),dotstyle); label("$F$", (-3.179956162676191,1.807637564594524), NE * labelscalefactor); dot((0.3574296255641460,1.389588409706417),dotstyle); label("$E$", (0.4188211874147766,1.486044695011927), N * labelscalefactor); dot((3.893636125792478,0.9936042076656184),dotstyle); label("$G$", (3.956342752823345,1.087882094576330), NE * labelscalefactor); dot((0.1820146103966589,2.872858546171026),dotstyle); label("$H$", (0.2503677795381781,2.971497473560114), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $\angle CDG = \angle DCE = \angle ABD.$ Similarly, $\angle DCG = \angle BAE.$ Hence, $\triangle ABE \sim \triangle DGC \implies \dfrac{AE}{CG} = \dfrac{BE}{DG} = \dfrac{AB}{CD}.$ Also, since $\angle FAB = \angle BCD$ and $\angle FBA = \angle FDC,$ $\triangle FAB \sim \triangle FDC \implies \dfrac{AB}{CD}= \dfrac{FB}{FD} = \dfrac{FA}{FC}$ (we could also get this by power of a point, but this is basically how PoP is proved). Combining them, $\dfrac{AE}{CG} = \dfrac{FA}{FC}.$ Now, note that $\angle FAC = 180^{\circ} - \angle BAD = 180^{\circ} - \angle ABC = \angle BCG,$ so $\triangle FAE \sim \triangle FCG.$ Similarly, $\triangle FBE \sim \triangle FDG.$ Finally, note that $\angle FHD = \angle DEF = 180^{\circ} - \angle BEF = 180^{\circ} - \angle FGD,$ so we're done. $\blacksquare$

This problem was nice and not too hard. First of all note that $\frac{FA}{AE}=\frac{FC}{CG} \Leftrightarrow \frac{FA}{AE}=\frac{FC}{ED} \Leftrightarrow \frac{FA}{FC}=\frac{AE}{ED}=\frac{AB}{CD}$ which is true since $\triangle{FAB} \sim \triangle{FCD}$.Also note that $\angle{FAE}=\angle{FAB}+\angle{BAE}=\angle{FCD}+\angle{EDC}=\angle{FCD}+\angle{DCG}=\angle{FCG}$.Combining these we get $\triangle{FAE} \sim \triangle{FCG}$.Thus $\angle{FHA}=\angle{FEA}=\angle{FGC}$.Also $\angle{AHD}=\angle{AED}=180-\angle{DEC}=180-\angle{DGC}$.Finally we get $\angle{FHD}=\angle{FHA}+\angle{AHD}=\angle{FGC}+180-\angle{CGD}=180-\angle{FGD}$ so $F,H,D,G$ are concyclic indeed.

Jeroen wrote: A solution requiring no additional points: We easily find $\triangle FAB \sim \triangle FCD$ and $\triangle EAB\sim \triangle GCD$. From these it follows that $\triangle FBE\sim\triangle FDG$. We now have $\angle FHD=\angle FED=180^\circ-\angle FEB=180^\circ-\angle FGD$, so $FHDG$ is cyclic. The best

This is very easy for IMO shortlist. My solution: Power of point $F$ and $E$ gets us that $\triangle AEF $ and $ \triangle FGC$ are similar (we also use the fact that $DGCE$ is a parallelogram). Now we have $\angle EFA=\angle GFC$ and because $ H$ is reflection of $E$ to $AD$ we have $\angle AFE=\angle AFH$ from which we get $ \angle HFG=\angle AFC$ and by trivial angle chase we get $ \angle HDG=180-\angle AFB \implies DHFG $cyclic. Done.

Here is mine. not much different still i thought to post. notice that $FAEB$ $~$ $FCDG$.(just angle chase). from here $\angle FGD$ = $\angle FEB$. this immediately gives the conclusion.

Lemma: Let $\triangle ABC$ be a triangle and let $M$ be the midpoint of $\overline{BC}.$ Then the locus $\mathcal{H}$ of points $P$ satisfying $\measuredangle ABP = \measuredangle PCA$ is a rectangular circumhyperbola of $\triangle ABC$ centered at $M.$ Proof: For any $P \in \mathcal{H}$, the isogonal conjugate $Q$ of $P$ verifies $\measuredangle QBC = \measuredangle BCQ.$ Thus, the locus of $Q$ is the perpendicular bisector $\tau$ of $\overline{BC}$, so $\mathcal{H}$ is the isogonal conjugate of $\tau.$ Since $\tau$ passes through the circumcenter of $\triangle ABC$, it follows that $\mathcal{H}$ is a rectangular circumhyperbola of $\triangle ABC.$ Meanwhile, if $P'$ is the reflection of $P$ in $M$, it is straightforward to check that $P' \in \mathcal{H}$ because $PBP'C$ is a parallelogram. Thus, $\mathcal{H}$ is symmetric about $M \implies M$ is the center of $\mathcal{H}.$ $\blacksquare$ Back to the problem at hand, let $\mathcal{H}$ be the rectangular circumhyperbola of $\triangle FCD$ centered at the midpoint $M$ of $\overline{CD}.$ Since $ABCD$ is cyclic, we have $\measuredangle FDE = \measuredangle ECF \implies E \in \mathcal{H}.$ But since $M$ lies on the nine-point circle $\gamma$ of $\triangle FDG$ (well-known), it follows that $E$ lies on the image $\gamma'$ of $\gamma$ under the homothety $\mathbf{H}(G, 2).$ But note that $\gamma'$ passes through $F, D$ and the orthocenter of $\triangle FDG.$ Thus, it is well-known that $\gamma'$ is just the reflection of $\odot(FDG)$ in $FD.$ Hence, the reflection $H$ of $E$ in $FD$ lies on $\odot(FDG)$, as desired. $\square$

Here's another proof of the key lemma (in a triangle $ABC$ with $DBCE$ a parallelogram and $BD,CD$ antiparallel, $AD,AE$ are isogonal): Let $P_1=BD\cap CE$ and $P_2=CD\cap BE$ be two points at infinity. By dual Desargues' involution theorem, on complete quadrilateral $DBCEP_1P_2$, $AB\to AC,AD\to AE, AP_1\to AP_2$ is an involution. But since $AP_1$ and $AP_2$ are isogonal, this involution is the reflection across the bisector of $\angle BAC$, and so $AD,AE$ are isogonal.

I just realized this was a G2. Then I solved the problem. Notice that $\angle{ADG} = 180^\circ - \angle{DAC} = 180^\circ - \angle{DBC} = \angle{FBE}$ because $AC // DG$, and see that $$\frac{DG}{DF} = \frac{EC}{DF} = \frac{BE}{BF}$$because $\frac{BE}{EC} = \frac{\sin \angle ECB}{\sin \angle DBC} = \frac{\sin \angle DBF}{\sin \angle FDB} = \frac{DF}{FB}.$ Thus triangles $DGF$ and $BEF$ are similar by SAS Similarity, so $\angle{DGF} = \angle{FEB}$. But $\angle{FEB} = 180^\circ - \angle{FED} = 180^\circ - \angle{FHD}$, so cyclicity is proved.

bary bash plagues me Note that $\angle FHD = \angle FH'D$ where $H'$ is the point that makes $EFH'D$ a parallelogram. We now use barycentric coordinates. Let our reference triangle be $FDC$, so that $F=(1,0,0)$, $D=(0,1,0)$, and $C=(0,0,1)$. Now let the circumcircle of $(ABCD)$ be $-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0$. Points $C$ and $D$ make $v=0$ and $w=0$. Thus the circumcircle is $-a^2yz-b^2xz-c^2xy+(x+y+z)(ux)=0$. Intersecting the circle with lines $FC$ and $FD$, we get $B=(b^2-w,0,w)$ and $A=(c^2-w,w,0)$. Thus $E=((b^2-w)(c^2-w),(b^2-w)w,(c^2-w)w)$. Now let $E=(p,q,r)$ in homogenized coordinates. Then $H'=(1-p,1-q,-r)$ and $G=(-p,1-q,1-r)$. We now look at the circles going through $F$ and $D$. They make it so that the equation is $-a^2yz-b^2xz-c^2xy+(x+y+z)(wz)=0 \equiv w=\frac{a^2yz+b^2xz+c^2xy}{z(x+y+z)}$. If we prove this value is the same for $H'$ and $G$ we'll be done. We have \[ \frac{a^2(1-q)(1-r)-b^2p(1-r)-c^2p(1-q)}{1-r} = \frac{-a^2r(1-q)-b^2r(1-p)+c^2(1-p)(1-q)}{-r} \equiv -a^2r(1-r)(1-q) + b^2r(1-r)p + c^2pr(1-q) = -a^2r(1-r)(1-q)-b^2r(1-r)(1-p) + c^2(1-p)(1-q)(1-r) \equiv \]\[ b^2r(1-r) = c^2(1-q)(1-p-r) = c^2q(1-q) \equiv b^2 \cdot \frac{w(c^2-w)}{b^2c^2-w^2} \cdot \frac{b^2c^2-w(c^2-w)-w^2}{b^2c^2-w^2} = c^2 \cdot \frac{w(b^2-w)}{b^2c^2-w^2} \cdot \frac{b^2c^2-w(b^2-w)}{b^2c^2-w^2} \equiv b^2w(c^2-w)c^2(b^2-w) = c^2w(b^2-w)b^2(c^2-w), \]which is obviously true so we're done.

Here's a cool solution using parallelogram isogonality lemma, which I also learned recently. It states that for a point P inside a triangle ABC with ABP=ACP, the point Q such that BPCQ is a parallelogram, AQ-AP are pairwise isogonal lines. Sketch. If we let D be the point s.t. APBD is a parallelogram, we have that BDQ is congruent to APC by a translation of vector AD. (This is just done by length equalities and parallel lines.) Now, from DQB=ACP=ABP=DAB, ADBQ is cyclic. Then BAQ=BRQ=PAC, as desired. $\square$ Now in the problem, from FDE=ACF (cyclic quad), by the lemma we have that DGF=AXF=ACF+CFG=FDB+DFE=180-FED=180-FHD, as desired. $\blacksquare$

By BrMO 2013/2, we have $\angle DFG=\angle CFE.$ Thus, we have $\angle DGF=180-\angle DFG-\angle FDG=180-\angle FDG-\angle CFE=\angle DAC-\angle CFE.$ Then $\angle DHF=\angle DEF=180-\angle ADB-\angle DFE=180-\angle DAC+\angle DFC-\angle DFE=180-\angle DAC+\angle CFE=180-\angle DGF,$ done.

OK! That's a G2. Not very hard unless I tripped? Let $G'$ be the reflection of $G$ across the midpoint of $FD$. Note that then $GFG'D$ is a parallelogram. Now, notice that we have $$FG'=GD=EC \text{ and } FG'\parallel GD \parallel EC$$and thus $FG'EC$ is in fact also a parallelogram. Now, this means $$\angle FG'E = \angle ECF = \angle EDF$$and thus $EDG'F$ must be cyclic. But now, note that $$\triangle DG'F \cong \triangle DGF \text{ and } \triangle EDF \cong \triangle {DHF}$$($H$ is the reflection of $E$ over AF). Thus, quadrilateral $HDGF$ is in fact congruent to quadrilateral $EDG'F$(The individual triangles fall on top of each other when reflecting over $AF$ as necessary). Thus, in turn $HDGF$ must also be cyclic, which was indeed the required conclusion.

My solution described in 1 sentence coz Im lazy to writeup entirely lol A lengthy though straightforward legnth and similar triangle chase helps us prove that $\triangle FBE \sim \triangle FDG$, which implies the result.

Note that since quadrilateral $AFBE \sim CFDG$. $$\angle FEB = \angle DGF = 180^{\circ}-\angle FED = 180^{\circ} - \angle DHF$$So opposite angles in quadrilateral $DHFG$ sum to $180^{\circ}$ implying the points are concyclic $\blacksquare$

We have $\triangle FDG \sim \triangle FBE$ from $\measuredangle FDG = \measuredangle FAC = \measuredangle EBF$ and $\frac{FD}{FB} = \frac{GD}{GE} = \frac{CD}{AB}$. Thus \[\measuredangle FGD = \measuredangle FEB = \measuredangle FED = \measuredangle FHD. \quad \blacksquare\]

We have $\triangle FDG \sim \triangle FBE$. We use SAS congruence; note that \[\angle FDG = 180^{\circ} - \angle DAC = 180^{\circ} - \angle DBC = \angle FBE.\]Furthermore, \begin{align*} & \frac{FB \cdot DG}{FD \cdot EB} \\ = ~ & \frac{FB \cdot EC}{FD \cdot EB} \\ = ~ & \frac{EC}{EB} \cdot \frac{FB}{FD} \\ = ~ & \frac{\sin \angle EBC}{\sin \angle ECB} \cdot \frac{\sin \angle FDB}{\sin \angle FBD} \\ = ~ & 1. \end{align*} As a result of $\triangle FDG \cong \triangle FBE$, we have \[\angle FGD = \angle FEB = 180^{\circ} - \angle FED = \angle FHD, \]hence $D, $H$, $F$ and $G$ are concyclic. edit: oooops why did i do this i am stupid

We have $\angle DGC = \angle DEC = \angle AEB$ and $\angle F = \angle F$ so $FAEB \sim FCGD$. Then angle chasing gives us $\angle DHF = \angle DEF = 180^{\circ} - \angle BEF = 180^{\circ} - \angle DGE$, so $GDHF$ is cyclic.

Claim: $\triangle FAE \sim \triangle FCG$ Proof: Since $\tfrac{FA}{FC} = \tfrac{AB}{DC} = \tfrac{AE}{DE} = \tfrac{AE}{CG}$ and $$\measuredangle FAE = \measuredangle DAC = \measuredangle DBC = \measuredangle BDC + \measuredangle DCB = \measuredangle GCD + \measuredangle DCB = \measuredangle GCB,$$the conclusion follows by SAS. Then the conclusion follows since $$\measuredangle FHD = \measuredangle DEF = \measuredangle DEA + \measuredangle AEF = \measuredangle DEC + \measuredangle FGC = \measuredangle CGD + \measuredangle FGC = \measuredangle FGD.$$

The reflection is arbitrary, so we get rid of point $H$ and instead note that it suffices to show $\angle FED + \angle FGD = 180$. First, we claim that lines $FE$ and $FG$ are isogonal with respect to $\angle CFD$. This can be shown using the parallelogram isogonality lemma. Alternatively, we can prove this by showing $\triangle FAE \sim \triangle FCG$ by SAS similarity. Indeed, using directed angles, \[ \angle FAE = \angle DAE = \angle DBC = \angle GCF, \]and furthermore, \[ \frac{FA}{FC} = \frac{AB}{CD} = \frac{AE}{ED} = \frac{AE}{CG}, \]so the claim is proven. Then, easy angle chasing yields $\angle FEB = \angle DGF$, and the desired result shortly follows.

From $\angle FDE = \angle FCE$ we have that $FE$ and $FG$ are isogonal wrt $\angle DFC$ (by a well known lemma). So we get \begin{align*} \angle FGD &= 180^\circ - \angle DFG -\angle FDC - \angle CDG\\ &= 180^\circ - \angle BFE - \angle FBA - \angle ABE\\ &= \angle FEB \end{align*} Also, from the reflection we have that $\angle FHD = \angle FED = 180^\circ - \angle FGD$. Done!

Suppose that $K$ is a point such as $FECK$ is parallelogram. Then $FK = EC = DG$ and $FK \parallel EC\parallel DG,$ so $FKGD$ is parallelogram. Hence $\angle{FKG} = \angle{FDG} = \angle{FDE} + \angle{EDG} = \angle{FCE} + \angle{ECG} = \angle{FCG}$ or $F, K, C, G$ lie on a circle. From this, we have $\angle{FGD} = \angle{GFK} = 180^{\circ} - \angle{KCG} = 180^{\circ} - \angle{FED} = 180^{\circ} - \angle{FHD}$ or $F, H, D, G$ lie on a circle

back from a very long break from doing math ok so we can just angle chase to get that $GCD$ and $EAB$ are similar and since $ABCD$ is cyclic $FBA$ and $FDC$ are similar too with literally the same exact common ratio then $GDC=EBA$ and $CDF=ABF$ so $CDF=EBF$ then $GDF$ and $EBF$ are similar so $BEF=DGF$ however $DEB$ is collinear so $BEF+DEF=180$ and since $H$ is the reflection of $E$ across $AD$ so $BEF+DHF=180$, but $BEF=DGF$ so $DGF+DHF=180$ thus $DHFG$ are concyclic

It suffices to show that \[180^{\circ}=\angle FHD+\angle FGD=\angle FED+\angle FGD=360^{\circ}-\angle DFE-\angle FDE-\angle GFD-\angle FDG.\]But note that \[\angle FDE+\angle FDG=\angle FDE+\angle CDG+\angle EDC+\angle FCE=180^{\circ}-\angle F,\]and $\angle DFE+\angle GFD=\angle F$ by the First Isogonality Lemma, so we are done. $\blacksquare$