Solution: Let $L, M$ be the intersection of $EF$ and $DC$ and $AB. GH$ cuts $EF$ at $J$. Applying Gauss's line we get $J$ is the midpoint of $EF$. Since $(FEML)=-1$ then $JE^2=JM.JL. (1)$ On the other side, $(ILDC)=-1$ and $H$ is the midpoint of $DC$ then $IL.IH=ID.IC$. Similarly, $IM.IG=IA.IB$ So $IM.IG=IL.IH$, which follows that $MGHL$ is a cyclic quadrilateral. Hence $JM.JL=JG.JH (2)$ From $(1)$ and $(2)$ we obtain $JE^2=JG.JH$. We are done.

Consider the transformation that takes $AB$ into $CD$ after a reflection on the internal bisector of $\angle AFB$ and a dilation of factor $\frac{FC}{FA}=\frac{FD}{FB}$ (equality holds because $ABCD$ is cyclic, hence $FA\cdot FD=FB\cdot FC$ is the power of $F$ with respect to its circumcircle). This transformation takes $G$ into $H$, and $E$ into $E'$ such that $CE'D$ is similar to $AEB$. But $AEB$ is also similar to $DEC$, or $CE'D$ and $DEC$ are equal, and $CEDE'$ is a parallelogram. Similarly, $E$ is the result of applying the transformation to $E''$ such that $AEBE''$ is a parallelogram. Now, $E',E''$ are on the symmetric to $FE$ with respect to the internal bisector of $\angle AFB$, or $P,E',E''$ are collinear. Note also that $G,H$ are the respective midpoints of $EE',EE''$, or $GH\parallel E'E''$. Now, the transformation preserves angles, or $\angle FEG=\angle FE'H=\angle E''E'E=\angle GHE$, and by semiinscribed angles, the conclusion follows.

April wrote: Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$. It was posted before here. Best regards, Majid.

As a matter of fact, exactly this problem (though with another designaions) had been proposed at an olympiad in St. Petersburg (10th grade, problem 7) several months before the shortlist was formed (14.12.2008). Can anyone comment on this coincidence, and how come that the problem from quite a well-known competition managed to pass all checks and had real chances to be posed to those who had been solving it before? Translation of the official wording from Russian: A circle passing through points $A$ and $C$ of a triangle $ABC$ intersects its sides $AB$ and $BC$ at points $Y$ and $X$ respectively. The segments $AX$ and $CY$ intersect at a point $O$. Let $M$ and $N$ be the midponits of the segments $AC$ and $XY$. Prove that the line $BO$ is tangent to the circumcircle of $MON$.

BlackMax wrote: As a matter of fact, exactly this problem (though with another designaions) had been proposed at an olympiad in St. Petersburg (10th grade, problem 7) several months before the shortlist was formed (14.12.2008). Can anyone comment on this coincidence, and how come that the problem from quite a well-known competition managed to pass all checks and had real chances to be posed to those who had been solving it before? I was in the Problem Selection Committee for the 2008 IMO in Madrid. We looked everywhere for precedents in competitions, magazines or online site (such as this one) of the longlisted problems. We scratched quite a few problems due to the fact that, either the exact problem, or a subtle variation which introduced no concept variation, had already been proposed/published/solved. Imagine our distress and shame when it was pointed out that still a couple of shortlisted problems (I don't remember exactly whether 2 or 3) had in fact been proposed in some competition; at least, these problems had not been given a wide exposure to the public and, if I remember correctly, no online source for the material could be quoted, just the booklet from the competition; nonetheless we learnt the hard way that it is close to impossible to check everything... there is just too much material out there (which is great for students preparing themselves for the competition, but not for submission of original problems). The only thing that can be done is, when you are on a PSC or Jury, do your best in order to find the problems that, through happenstance or just plain bad will (I will always prefere to believe in the first case if I do not have enough evidence against - innocent until proven guilty), slipped into the longlist/shorlist. And of course, encourage fair play with your own example... There is something else that also distresses me quite a bit: shortlists are supposed to be kept secret until the following year's IMO, but we find here and there these problems breaking out into the open. I think shortlist problems are a very valuable material for student training/selection (heck, I use it!), but whenever such a problem is used, it should be made sure that the students who are exposed to them, know very well that it is "top-secret" stuff and should not be discussed or commented outside of the team... And just for the sake of extra security, I only work with students on the previous year IMO shortlist about 1-2 weeks before the following IMO...

livetolove212 wrote: $(FEML)=-1$ How do we know that M and L are conjugate points? Is this just a well-known fact?

Sorry to revive this topic, but I suddenly felt the urge to post the solution to this problem that I came up with during a test at MOP. WLOG let $E,F$ be on opposite sides of $AB$, and let $E$ be closer to $BC$ than to $AD$. Let line $EF$ intersect lines $AB$ and $CD$ at $J$ and $K$, respectively. Now, if we can prove that $\frac{\sin \angle GHE}{\sin \angle HGE}=\frac{\sin\angle GEJ}{\sin\angle HEK}$, then we are done, since for fixed $0<\alpha<\pi$, $\frac{\sin x}{\sin( \alpha-x)}$ is increasing, so the above would imply that $\angle GHE=\angle GEJ$, or that $EF$ is tangent to the circumcircle of $GEH$. Thus we will now try to prove that \[\frac{\sin \angle GHE}{\sin \angle HGE}=\frac{\sin\angle GEJ}{\sin\angle HEK}\] We know that $\Delta FAB$ is similar to $\Delta FCD$, $\Delta FAC$ is similar to $\Delta FBD$, and $\Delta AED$ is similar to $\Delta BEC$. Also, By Ceva's theorem, \[\frac{GJ}{JB}=\frac12\left(\frac{AJ}{JB}-1\right)=\frac12\left(\frac{CF\cdot DA}{BC\cdot FD}-1\right)=\frac{CF\cdot DA-BC\dot FD}{2BC\cdot FD}\] Similarly, \[\frac{HK}{KC}=\frac{AD\cdot BF - FA\cdot CB}{2FA\cdot CB}\] Now we have: \begin{eqnarray} 1&=&\left(\frac{FB}{FD}\right)^2\left(\frac{FD}{FB}\right)^2\\ &=&\frac{FB^2-FA^2}{FD^2-FC^2}\cdot\frac{FB}{FD}\cdot\left(\frac{FD}{FB}\right)^3\\ &=&\frac{FA(FD-FA)-FB(FC-FB)}{FD(FD-FA)-FC(FC-FB)}\cdot\frac{FB}{FD}\cdot\left(\frac{CD}{AB}\right)^3\\ &=&\frac{FA\cdot DA-BC\cdot FB}{AD\cdot AF-FC\cdot CB}\cdot\frac{FB}{FD}\cdot\left(\frac{CD}{AB}\right)^3\\ &=&\frac{\frac{CD}{AB}\left(FA\cdot DA-BC\cdot FB\right)}{\frac{AB}{CD}\left(AD\cdot AF-FC\cdot CB\right)}\cdot\frac{FB}{FD}\cdot\frac{CD}{AB}\\ &=&\frac{CF\cdot DA-BC\cdot FD}{AD\cdot BF-FA\cdot CB}\cdot\frac{FA}{FD}\cdot\frac{FB}{FA}\cdot\frac{CD}{AB}\\ &=&\frac{\frac{CF\cdot DA-BC\cdot FD}{2BC\cdot FB}}{\frac{AD\cdot BF-FA\cdot CB}{2FA\cdot BC}}\cdot\frac{FB}{FA}\cdot\frac{CD}{AB}\\ &=&\frac{\frac{JG}{BJ}}{\frac{KH}{CK}}\cdot\frac{FB}{FA}\cdot\frac{CD}{AB}\\ &=&\frac{BE}{CE}\left(\frac{CD}{AB}\right)^2\cdot\frac{\frac{JG}{BJ}}{\frac{KH}{CK}}\cdot\frac{\frac{FB\sin\angle FBE}{EF}}{\frac{FA\sin\angle FAE}{EF}}\\ &=&\frac{BE}{CE}\left(\frac{CD}{AB}\right)^2\cdot\frac{\frac{JG}{BJ}}{\frac{KH}{CK}}\cdot\frac{\sin\angle JEB}{\sin\angle AEJ}\\ &=&\left(\frac{CD}{AB}\right)^2\cdot\frac{JG}{KH}\cdot\frac{\frac{BE\sin\angle JEB}{BJ}}{\frac{CE\sin\angle KEC}{CK}}\\ &=&\left(\frac{CD}{AB}\right)^2\cdot\frac{JG}{KH}\cdot\frac{\sin\angle BJE}{\sin\angle CKE}\\ &=&\left(\frac{CD}{AB}\right)^2\cdot\frac{JG}{KH}\cdot\frac{\sin\angle GJE}{\sin\angle HKE}\\ &=&\left(\frac{CD}{AB}\cdot\frac{GE}{HE}\right)^2\cdot\frac{HE}{GE}\cdot\frac{\frac{JG\sin\angle GJE}{GE}}{\frac{KH\sin\angle HKE}{HE}}\\ &=&\frac{\frac{\sin\angle GEJ}{\sin\angle HEK}}{\frac{\sin\angle GHE}{\sin\angle HGE}} \end{eqnarray} so we are done. Notes: steps 9,10,11,12,14, and 15 use the law of sines; step 9 follows from the fact that $\Delta AED$ is similar to $\Delta BEC$; step 15 followes from the fact that $HE$ and $GE$ are the corresponding medians in similar triangles $\Delta DEC$ and $\Delta AEB$.

Take the relation (4) from http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=202546&hilit=Ptolemy and apply it to the quadrilateral $DECF$ and get $\frac{GJ}{HJ}=\left(\frac{GE}{HE}\right)^2$, i.e. $FE$ is tangent to circle $(EGH)$. Here $J$ is the midpoint of $EF$ and, as per Gauss line theorem $G, E$ and $J$ are collinear. Best regards, sunken rock

mgao wrote: How do we know that M and L are conjugate points? Is this just a well-known fact? Let $IE$ hit $BC$ at $X$. $-1 = (F,X; B,C) = I(F,X; B,C) = I(F, E; M, L) = (F, E; M, L)$.

The subject problem has been proposed at a Romanian contest as well; its outstanding official solution can be read here: http://forum.gil.ro/viewtopic.php?f=25&t=255&p=400#p400 . I hope none of you will face problems in understanding; if there is the case, I might translate it. Best regards, sunken rock

sunken rock wrote: The subject problem has been proposed at a Romanian contest as well; its outstanding official solution can be read here: http://forum.gil.ro/viewtopic.php?f=25&t=255&p=400#p400 . I hope none of you will face problems in understanding; if there is the case, I might translate it. Best regards, sunken rock Very beautiful solution indeed, thanks for the link!

This is my solution, I rather like it. Let $O$ be the circumcenter of $ABCD$. Let the intersection of $AB$ and $CD$ be $X$. Lemma: The polar of $X$ with respect to the circumcircle of $ABCD$ is $FE$. Proof: Consider $G'$ and $H'$, the poles of $AB$ and $CD$ respectively, which lie on the polar of $X$. By Pascal's Theorem on $AACBBD$ and $CCADDB$, we have that $E, F, G', H'$ are collinear. We invert around $O$. Let the images of $A,B,C,D,E,F,G,H,X$ be $A', B'...$and so on, respectively. We want to show that the circumcircles of $XOE'$ and $E'G'H'$ are tangent, since inversion preserve tangency. Let the foot of the perpendicular from E' to XO be Z. Let the pole of $E'Z$ be $Y$. Let $E'Y$ intersect $CD$ at $K$, whose image after inversion about $O$ is $K'$. The circumcenter $K'$ of $E'G'H'$ is the intersection of $CD$ and the perpendicular bisector of $E'H'$. Let $Y'$ be the intersection of $EH'$ and $ZO$. $EKHO$ and $E'ZK'O$ are cyclic because $\angle{OEK}, \angle{OHK}, \angle{E'ZO}, \angle{E'K'O}$ are right, and thus, $\angle{EYO} = \angle{EKO} + \angle{YOK}=\angle{EHO} + \angle{ZE'Y'} =\angle{OE'Y'} + \angle{ZE'Y'}=\angle{EE'Z}=\angle{EY'O}$ Thus $Y'=Y$, and so $E', Y, K'$ collinear $\Rightarrow E', Z, K$ collinear. Since $Z,K$ are the circumcenters of $XOE'$ and $E'G'H'$, and $E'$ is a common point, $E'$ is their point of tangency, and we are done.

New points: $O$ center of the circle $(ABCD)$, $X,Y$ intersections of tangents in $C, D$ and $A,B$ on $(ABCD)$, $N, M$ intersections of $XY$ with $CD$ and $AB$, $P$ interesection of $CD$ and $AB$, $T$ intersection $XY$ and $OP$. It is known that $XY$ passes through $E$ and that $XY$ is polar of $P$ with respect to $(ABCD)$. We will show that $HNMG$ is cyclic. This is enough because then $ \angle HEN= \angle ENC - \angle EHN= \angle HGM - \angle EGM = \angle HGE $. Now we have $ \angle OGM=\frac {\pi}{2}= \angle OTM$ so $OTMG$ is cyclic. Similar $OTNH$ is cyclic. So we have $PM \cdot PG=PT \cdot PO=PN \cdot PH$; $PM \cdot PG=PN \cdot PH$ and we have $HNMG$ is cyclic.

I was just wondering about this problem and tried to solve it with spiral symmetries; it revealed many facts about spiral centres. Here is the solution .. Consider the spiral symmetry with centre $ S $ which sends $ A $ to $ D $ , $ B $ to $ C $ . $ GE \cap CD = H' $ and $ HE \cap AB = G' $, $ AB \cap CD = I $ then. It is well-known that $ S $ lies on polar $ e = IF$ of $ E $ and is the inverse of $ E$ in circumcircle of $ ABCD $ . Clearly, $ S \in $ the circle $ \gamma $ passing through $ E $ , $ B $ and $ A $. $ \textit{Lemma 1.} $ The tangent $ t $ to the circle $ \gamma $ at $ S $ and the line $ l $ parallel though $ CD $, passing through $ E $ intersect at $ AB$. $ \textit{Proof:} $ The line $ l $ is tangent to the circumcircle $ \omega $ of $ \triangle AEB $. Let $ l\cap t = U $. Line $ AB $ is the radical axis of circles $ \gamma $ and $ \omega_{1} $ so it's enough to show that $ US = UE $. Then, $ \angle ESU = ESI - \angle USI = \angle FSU - 90 = \angle FSA + \angle ASU - 90 =$ $ \angle CBA + \angle AFS - 90 = \angle IDA + AFS -90 = \angle DFI + \angle IDF - 90 = 90 - \angle FID $. We use oriented angles to avoid case work. Also, $ \angle UES = \angle IVS = 180 - \angle VSI - \angle SIV = 90 - \angle SIB = 90 - \angle FID $ and this proves the lemma. $ \textit{ Lemma 2. } $ Lines $ SG' $ and $ FE $ intersect at $ E' $ , $ E'$ on the circle $ \gamma $ . $ \textit { Proof : } $ Let $ FE \cap \gamma = E'$. The bundle $ ( FI , FE ; FA , FB ) $ is harmonic and so is the quadrilateral $ S E' A B$. For triangle $ BEA $, lines $ EG'$ and $ EU$ are respectively, the symmedian and tangent to $ \omega $ at $ E $, so the range $ ( U , G' ; A , B ) $ is harmonic and so is the bundle $ ( SU , SG' ; SA , SB )$ and so intersecting this bundle with $ \gamma $ gives a harmonic quadrilateral $ S G'' A B $ and so $ G'' = E' $ which proves the lemma. We now prove the main result. Let $ H'G' $ intersect $ CB $ in $ X$ . Since the initial spiral symmetry also sends $ G'$ to $ H'$, the four points $ X$ , $ B$ , $ G' $ and $ S $ lie on a circle, so $ \angle G'SB = G'XB $. As, $ SE'BF $ is cyclic, $ \angle G'SB = \angle E'SB = \angle E'FB = \angle EFB $. So, $ \angle G X ' B = \angle EFB $ which implies that the antiparallel to $ GH $ in triangle $ HEG $ is parallel to $ EF $ which completes the proof.

Let the point H' such that CFDH' is a parallelogram, and G' be a point such that FBG'A is a parallelogram. Extende FE to intersect CD at J and AB at I. Note that quadrilaterals ECFD and EBG'A are similar since ▲BFA~▲CFD, with F corresponding to G'. Since EF correspands to EG', EH' corresponds to EF, and ∠FEH'=∠G'EF (from similar quadrilaterals), points E, H', G' are collinear. Let EG' intersect CD at K and AB at L. ▲EKD~▲EIB (corresponding parts of quadrilaterals), thus JKLI is cyclic. Since HG || H'G', JHGI is cyclic. Then, ∠HGA=∠FJH and ∠FGB=∠JHF since FH and FG are medians of similar triangles BFA and CFD. We have ∠HGA+∠FGB+∠FGH=180=∠FJH+∠JHF+∠JFH. Using the above angle equalities, ∠JFH= ∠FGH. Thus EF is tangent to the circumcircle of FGH.

Nice problem.

Note Newton Gauss line, so $M$ be the midpoint of $EF$, and $M \in GH$. Also, let $O$ be circumcentre of $ABCD$ and $AB, CD$ meet at $I$ where it follows $OGHI$ are concyclic, or $GH, EF$ are antiparallel, or $EF \cap AB = X, EF \cap CD = Y, G, H$ are concyclic. But $ME^2 = MX \cdot MY = MG \cdot MH$, so the problem just got destroyed.

sunken rock wrote: The subject problem has been proposed at a Romanian contest as well; its outstanding official solution can be read here: http://forum.gil.ro/viewtopic.php?f=25&t=255&p=400#p400 . I hope none of you will face problems in understanding; if there is the case, I might translate it. Best regards, sunken rock can you tell me why M is midpoint of AO? the rest is clear

@Blitzkrieg97: $AEODBC$ is a complete quadrilateral ($AEOB$ the quad, and $B,C$ the intersections of opposite sides), hence $AO, DE, BC$ are its diagonals, which have their midpoints collinear (Newton-Gauss line). Best regards, sunken rock

We want to prove that $\angle FEG = \angle EHG$, which is equivlent to \[\frac{(e-g)(e-h)}{(g-h)(e-f)} \in \mathbb{R}\]First, notice by Brocard's, we have $\overline{EF} \perp \overline{OP}$ where $P = \overline{AB} \cap \overline{CD}$. Thus since $EF$ is just $OP$ rotated $90^{\circ}$ and scaled, we have: \[e-f = k_1 \cdot i \cdot p = k_1 \cdot i \cdot \frac{ab(c+d)-cd(a+b)}{ab-cd}\]for some real constant $k_1$. Next, by SSS similarity, $\Delta BEG \sim \Delta CEH$, so \[\frac{(h-e)(g-e)}{(b-e)(c-e)} \in \mathbb{R} \implies (h-e)(g-e) = k_2 \cdot (b-d)(c-a)\]for some real constant $k_2$. Lastly, we see that $g-h = \frac{a+b-c-d}{2}$, so our desired expression is \[i \cdot \frac{ab(c+d)-cd(a+b)}{ab-cd} \cdot (a+b-c-d) \cdot \frac{1}{(b-d)(c-a)} \in \mathbb{R}\]which can be easily verified by taking the conjugate to see that they're equal.

Define $I=EF\cap GH$, $J=EF\cap AB$, $K=EF\cap CD$, $L=DJ\cap BC$, $M=AK\cap BC$ and $N=AB\cap CD$. Note that since $\overline{DB},\overline{CA},\overline{FK}$ intersect at $E$, then from Ceva-Menelaus' we get that $(F,C;L,K)=-1$. Project this onto line $\overline{FK}$ with perspective at $D$ to get $(F,E;J,K)=-1$. From Newton-Gauss Lemma, we have that $I$ is the mid-point of segment $\overline{EF}$. Sinc $(F,E;J,K)=-1$, we have that $IF^2=IJ\cdot IK$. We need to show that $IF^2=IG\cdot IH$, so we need to show that $G,H,K,J$ are cyclic. Note that since $\overline{DL},\overline{FE},\overline{BA}$ concur at $J$, we have that $(F,B;L,C)=-1$ (from Ceva-Menelaus'). Taking perspective at $D$, we get that $(A,J;B,N)=-1$. So, under inversion about $G$ with radius $\overline{GA}$, $J$ goes to $N$. We have \[NJ\cdot NG=NG^2-NG\cdot JG=NG^2-GB^2=NA\cdot NB.\]Similarly, $NC\cdot ND=NK\cdot NH$. However, since $ABCD$ is cyclic quadrilateral, $NA\cdot NB=NC\cdot ND$, which gives $NJ\cdot NG=NK\cdot NH$, which means $G,H,K,J$ are concyclic. $\blacksquare$

why is 2016 usamo 5 hard Let $O$ be the circumcenter of $ABCD$. Let $AB\cap CD=X$. WLOG assume that $F$ lies on rays $DA$ and $CB$ and that $X$ lies on rays $AB$ and $DC$. With this configuration, it suffices to prove something like \[\angle BDC+\angle BDC+\angle DEC=90^{\circ}-\angle AXO+90^{\circ}-\angle OXD=180^{\circ}-\angle AXD.\](We get this by noting $\angle(FE,DC)+\angle(GH,DC)=2\angle(\ell,DC)$ where $\ell$ is the angle bisector of $\angle DEC$. Then we do some angle chasing.) Now: \[\angle BDC+\angle AXD+\angle BDC+\angle DEC=180^{\circ}-\angle ACD+\angle AXD+\angle BDC=180^{\circ}.\]Done. $\blacksquare$ post #2 is interesting, using cevian harmonic and midpoint pop spam and stuff to get result (trick there is gauss line i guess, did not think of that at all oops. need to like stop spamming and memorize configs :nerd: though ig geo is dead)

Let $M$ be the midpoint of $\overline{EF}$. The problem is equivalent to showing $ME^2 = MG \cdot MH$. First, assume that $\angle CAD = \angle CBA = 90^{\circ}$. In this case, the problem is true by the three tangents lemma; since $\angle HGB = \angle MBH = 90^{\circ}$, we have $MG \cdot MH = MB^2 = ME^2$.

Let $M$ be the midpoint of $EF$, it is well-known that $G,H,M$ are collinear. Let $X=AB \cap CD$ be the pole of $EF$, and let $P=EF \cap AB, Q=EF \cap CD$. Then $(X,P;A,B)=(X,Q;C,D)=-1$ and so $XP \cdot XG = XA \cdot AB = XC \cdot XD = XQ \cdot XH \implies G,H,P,Q$ are concyclic. Also note that $(E,F;P,Q)=-1$, so $ME^2=MP \cdot MQ = MG \cdot MH \implies ME$ is tangent to $(EGH)$, so we are done. $\square$

Denote $EF \cap AB = I$, $AB \cap CD = J$, $EF \cap DC = K$. By cevians (J, I; A, B) = -1. Also (J, I; A, B) = (J, K; C, D) = -1. From problem 1 (or G and H being midpoints of AB and DC respectfully) we get JA.JB = JI.JG and JC.JD = JK.JH. But from ABCD being cyclic we get that JA.JB = JC.JD $\Rightarrow$ we get that JI.JG = JK.JH $\Rightarrow$ IKHG is cyclic. Denote $GH \cap EF = M$. From Newton-Gauss line for ECFD it follows that M is midpoint of EF $\Rightarrow$ $ME^2 = MK.MI$. From IKHG cyclic we get MH.MG = MK.MI $\Rightarrow$ $ME^2 = MH.MG$ $\Rightarrow$ ME is tangent to (EHG) in E. We are ready.

Let $AB$ and $CD$ meet at $T$. The main point of this problem is to understand the directions of the lines $EF$, $GE$, and $HE$. $EF$ is not too hard since we know by Brocard that it's the polar of $T$, so it is perpendicular to $OT$. Now, using complex numbers, the tangency condition is $$\frac{(h-g)(f-e)}{(e-g)(h-e)}$$real. By the Brocard argument, we can replace $f-e$ with $t$ and instead show it is imaginary. Furthermore, since $\triangle EGB\sim\triangle EDC$, we can also replace $(e-g)(h-e)$ with $(g-b)(h-c)$, but we can also replace this with $(a-b)(d-c)$. Thus, it suffices to show that $$\frac{(c+d-a-b)t}{(a-b)(d-c)}$$is imaginary where $$t=\frac{ab(c+d)-cd(a+b)}{ab-cd},$$but this is clearly true as this becomes $$\frac{(c+d-a-b)(ab(c+d)-cd(a+b))}{(d-c)(a-b)(ab-cd)},$$but if we conjugate and multiply by $a^2b^2c^2d^2$, each factor on the bottom flips signs and the two top factors swap, so it becomes the negative, done. remark: The main idea of this solution is the similarity used to replace $(e-g)(e-h)$ with $(h-c)(g-b)$. The idea here is that although the tangency condition involves the directions of lines $GE$ and $HE$, it actually only incorporates the "sum" of the directions. The similar triangles allow us to move this sum to somewhere more simple. Alternative with harmonic sketch: let $GH$ intersect $EF$ at its midpoint $M$. then we wish to show that $MH\cdot MG=ME^2$, but by harmonic bundle we also have $MI\cdot MJ=ME^2$, where $I$ and $J$ are the intersections of $EF$ with $CD$ and $AB$. Thus, it suffices to show that $IJGH$ is cyclic, which follows from both $(TJ;BA)$ and $(TI;CD)$ being harmonic and applying power of a point.

April wrote: Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$. Proposed by David Monk, United Kingdom Let $AB\cap DC=K,FE\cap AB=I,EF\cap DC=J$ and $N$ the midpoint of $FE$. we have that $(A,B/I,K)=-1=(D,C/J,K)$ so since $G,H$ are midpoints we have that: $KG\cdot KI=KA\cdot KB=KC\cdot KD=KJ\cdot KH\Rightarrow G,I,J,H$ are cyclic. By Neton-Gauss line we have that $H,G,N$ are collinear. Now since $(J,I/E,F)=-1$ and $N$ is the midpoint we have that $NE^2=NI\cdot NJ=NG\cdot NH$ and we are done.

Let $\overline{FE}$ intersect $\overline{DC}$ and $\overline{BA}$ at $K$ and $L$ respectively. Let $\overline{GH} \cap \overline{FE}$, at $I$. Becuause of the existence of the Gauss Line $I$ is the midpoint of $EF$. Let $\overline{FE}$ intersect $CD$ and $AB$ at $K$ and $L$ respectively. Since $-1= (CD; JK) \stackrel{F} = (BA; JL)$ then by the midpoint lemma $JL \cdot JG = JA \cdot JB = JD \cdot JC = JK \cdot JH$, which implies $LKHG$ is cyclic by PoP. Now since $-1=(FE; LK)$ by the midpoint lemma $IE^2= IE \cdot IF = IL \cdot IK= IG \cdot IH$ which implies the tangency.

2009 G4 Let $FE$ intersect $AB$ and $CD$ at $X$ and $Y$. And let $HG$ intersect $FE$ at $K$. Let $AB$ intersect $CD$ at $L$. First of all, it’s well known that $K$ is the midpoint of $FE$, Also we know that $(CD;JL)=-1=(BA;IL)$. By PoP we know that $XYHG$ are cyclic. We also have that $(FE;XY)=-1$, hence by PoP and midpoint lemma we can show the desired tangency.

Let $M$ be the midpoint of $EF$, $EF \cap AB=X$ , $EF \cap CD =Y$ and $AB \cap CD=K$. Due to the existence of the Newton-Gauss Line, $\overline {G-H-M}$ are collinear. Also notice that the bundles $ (EF;XY),(AB;KX)$ and $(CD;KY)$ are all harmonic due to Ceva-Menelaus and projecting. Notice that since $M$ is midpoint of $EF$ ; $ME^2=MX.MY$ due to a well known property of harmonic bundles. Also since $G$ is midpoint of $AB$ $KA.KB=KG.KX$ due to a similar property as above. Analogously, $KC.KD=KY.KH$ but since $KA.KB=KC.KD$ due to Power of Point $\implies KG.KX=KH.KY$ so $G,X,Y,H$ are concylic $$\implies ME^2=MX.MY=MG.MH$$which shows that $ME$ which is $EF$ is tangent to $(EGH)$ as desired.

Let the intersections of $EF$ with $AB$ and $CD$ be $G_1$ and $H_1$. Let $AB \cap CD = K$. Let the midpoint of $EF$ be $N$. By Newton-Gauss we have that $G-H-M$. By Brokard's we have that $EF$ is the polar of $K$ wrt $(ABCD)$, so $(A, B; G_1, K) = (D, C; H_1, K) = -1$. By a well known lemma this implies that $KB \cdot KA = KG_1 \cdot KG$ and $KC \cdot KD = KH \cdot KH_1$. However by PoP we have $KB \cdot KA = KC \cdot KD$ so $GG_1H_1H$ is cyclic. By Brokard's again we have that $KE$ is the polar of $F$ so $(A, D; KE \cap AD, F) = -1$ and projecting from $K$ onto $EF$ we get that $(G_1, H_1; E, F) = -1$. This implies that $G_1$ and $H_1$ are inverses wrt $(EF)$. By PoP we get that $NH^2 = NG_1 \cdot NH_1 = NG \cdot NH$ which implies that $EF$ is tangent to $(EGH)$ as desired.

Different solution i guess. Let $E'$ and $E"$ be points such that $AEBE'$ and $DECE"$ be parallelograms. Let $\angle DCE"=\angle EDC=\angle BAC=\angle E'BA=\alpha$, $\angle ACD=\angle ABE=\angle E'AB=\angle E''DC=\beta$ and, $\angle ADB=\angle ACB=\theta$. From angle chasing $\angle FAE'=\angle FBE'=\theta$. From simple lemma $\angle EFA=\angle E'FB=\gamma$. Similarly, for $\triangle FDC$. We deduce that $F-E'-E"$. Since, $G$ is midpoint of $EE'$ and $H$ is midpoint of $EE''$ $\implies$ $FE''$ is parallel to $GH$. Thus, if $FE$ tangents to $\triangle EGH$, then, $FE$ tangents to $\triangle FEE"$. So, it is similar to prove $FE \cdot FE=FE^2$. LoS to $\triangle FAE$ and $\triangle FAE'$. $\frac {FE'}{\sin{\theta}}=\frac{FA}{\sin{(\alpha+\beta+\theta+\gamma)}}$ and $\frac {FE}{\sin{(\alpha+\beta+\theta)}}=\frac{FA}{\sin{(\alpha+\beta+\theta+\gamma)}}$. So, $\frac{FE'}{\sin{\theta}}=\frac{FE}{\sin{(\alpha+\beta+\theta)}}$. LoS to $\triangle FEC$ and $\triangle FE"C$. $\frac{FC}{\sin{(\alpha+\beta+\theta+\gamma)}}=\frac{FE}{\sin{\theta}}$ and $\frac{FC}{\sin{(\alpha+\beta+\theta+\gamma)}}=\frac{FE"}{\sin{(\alpha+\beta+\theta+\gamma)}}$. So, $\frac{FE}{\sin{\theta}}=\frac{FE"}{\sin{(\alpha+\beta+\theta+\gamma)}}$.Thus, by multiplication we are done.

Reflect $E$ over $G$, $H$ to points $K$, $L$. It suffices to show $EF$ is tangent to $(EKL)$. The first isogonality lemma tells us $FKL$ collinear is isogonal to $FE$ with respect to $\angle AFB$. Notice that quadrilaterals $FKEB$ and $FELD$ can be dissected along diagonal $BK$ and $DE$ into the similar triangles shown below, making the quadrilaterals themselves similar. However, we can then dissect along diagonals $FL$ and $FE$, giving the desired $\triangle FKE \sim \triangle FEL$. $\blacksquare$ [asy][asy] size(270); defaultpen(linewidth(0.4)+fontsize(8)); pair A, B, C, D, E, F, K, L; A = dir(95); B = dir(55); C = dir(350); D = dir(190); E = extension(A, C, B, D); F = extension(A, D, B, C); K = A+B-E; L = C+D-E; draw(A--B--C--D--F--B^^A--E--B--K--cycle^^C--E--D--L--cycle^^K--E--L--F--E^^unitcircle); draw(E--(E+.98*(K-E))--(E+.98*(B-E))--cycle^^L--(L+.99*(E-L))--(L+.99*(D-L))--cycle, red+linewidth(2)); draw(F--(F+.96*(K-F))--(F+.96*(B-F))--cycle^^F--(F+.99*(E-F))--(F+.99*(D-F))--cycle, blue+linewidth(2)); dot("$A$", A, dir(120)); dot("$B$", B, dir(40)); dot("$C$", C, dir(330)); dot("$D$", D, dir(210)); dot("$E$", E, dir(315)); dot("$F$", F, dir(90)); dot("$K$", K, dir(130)); dot("$L$", L, dir(270)); [/asy][/asy]

Denote $X, Y = EF \cap AB, CD$, $M$ as the midpoint of $EF$, and $J = AB \cap CD$. Then we have the harmonic bundles $(AB;XJ)$ and $(CD;YJ)$, which give \[JG \cdot JX = JA \cdot JB = JC \cdot JD = JH \cdot JY,\] so $GHXY$ is cyclic. We finish with the bundle $(EF;XY)$, as \[ME^2 = MX \cdot MY = MG \cdot MH. \quad \blacksquare\]