The fact that $ BCDE$ is concyclic is superfluous (it may make the problem easier, but it is not necessary for the problem to be true). Let $ N_{1}$ be the point on where $ LH$ intersects the circle (apart from $ H$). By using Pascal, with $ A' = A$ being a degenerate point, we obtain that the following points are collinear: $ AB \cap HC, AA' \cap HN, CA' \cap BN$ - this means $ DE, AC$ and $ BN_{1}$ are concurrent (since $ E = AB \cap HC, L = AA' \cap HN$), so if $ X$ is where $ AC, DE$ meet, then $ N_{1}$ is where $ XB$ intersects the circle again. If we similarly define $ N_{2}$ to be the point where $ MG$ intersects the circle, we will obtain, by similarly applying Pascal, that $ N_{2}$ is also where $ BX$ intersects the circle - this means $ N_{1} = N_{2}$, which solves our problem. In the original problem (with $ BCDE$ concylic), this gives us the cute result, that $ N$ is the Miquel point of $ BCDE$.

mathpk wrote: Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$. See all problems from APMO 2008 here, http://www.mathlinks.ro/viewtopic.php?p=1073977#1073977 If $ LH$ cut the circle at a X (different from H ) then $ B,X,E,D$ is cyclic . It mean that $ LH,MG$ cut on at a point on the circle.

Dear Mathlinkers, a want only to detail the interesting last solution : 1 let U be the meetpoint of BX and AC, Ta, Tc the resp. tangent to the circumcircle of ABC at A, C, and X the second meetpoint of LH with the circumcircle of ABC. 2. ULE is the Pascal's line of BXHCA Ta B 3. U, L, E and D are collinear 4. By a converse of Monge (or d'Alembert)'s theorem, B, X, E and D are concyclic 5. By a converse of Pascal's theorem applied to BXGAC Tc B, we are done... Sincerely Jean-Louis

mathpk wrote: Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$. See all problems from APMO 2008 here, http://www.mathlinks.ro/viewtopic.php?p=1073977#1073977 Let $ LH\cap (O) = \{K\}$ We have $ \angle LKA = \angle HCA = \angle LDA$ then $ ALKD$ is cyclic. Let $ ED\cap AC = \{R\}, RB\cap (O) = \{K'\}$ We have $ RK'.RB = RA.RC = RE.RD$ so $ K'EDB$ is cyclic. $ \Rightarrow \angle K'DE = \angle K'BE = \angle K'AL$ so $ ALK'D$ is cyclic Therefore $ K'\equiv K$ we get $ KEDB$ is cyclic. $ \Rightarrow \angle AKC = \angle EBD = \angle EKD$ But $ \angle KDE = \angle KAL = \angle KCA$ then $ \Delta KED$~$ \Delta KAC$ $ \Rightarrow \angle KED = \angle KAC = \angle KCM \Rightarrow KECM$ is cyclic. Let $ MG\cap (O) = \{K''\}$ then similarly, we obtain $ K''ECM$ is cyclic so $ K''\equiv K$->QED

Drop the condition that $AEDC$ is cyclic (this is the hardest part). Now let $P = \overline{AD} \cap \overline{CE}$. Take a projective transformation which preserves $(ABC)$ and sends $\overline{BP} \cap \overline{AC}$ to the center of the circle. Then it's plain by Ceva that $\overline{ED} \parallel \overline{AC}$, so $ALCM$ is a rectangle. Thus $ALHE$ and $CMGD$ are concyclic, and then straight angle chasing.

Let $LH\cap {\Gamma}=K$, the tangent lines of $\Gamma$ at $A$ and $C$ be $t_1$ and $t_2$, $KG\cap {t_2}=M'$ and $BK\cap AC=P$, and the line $ELDM$ be $l$. Pascal's Theorem implies ${t_1}\cap KH=L$, $AB\cap HC=E$, $BK\cap CA=P$ are collinear i.e. $P$ lies on $l$. Pascal's Theorem implies ${t_2}\cap KG=M'$, $CB\cap GA=D$, ${BK\cap AC}=P$ are collinear i.e. $M'$ lies on $l$ i.e. $t_2, KG, l$ concurrent. Since ${t_2}\cap l=M$, $M,G,K$ are collinear i.e. $LH\cap MG=K$ i.e. on ${\Gamma}$.

epitomy01 wrote: The fact that $ BCDE$ is concyclic is superfluous (it may make the problem easier, but it is not necessary for the problem to be true). Let $ N_{1}$ be the point on where $ LH$ intersects the circle (apart from $ H$). By using Pascal, with $ A' = A$ being a degenerate point, we obtain that the following points are collinear: $ AB \cap HC, AA' \cap HN, CA' \cap BN$ - this means $ DE, AC$ and $ BN_{1}$ are concurrent (since $ E = AB \cap HC, L = AA' \cap HN$), so if $ X$ is where $ AC, DE$ meet, then $ N_{1}$ is where $ XB$ intersects the circle again. If we similarly define $ N_{2}$ to be the point where $ MG$ intersects the circle, we will obtain, by similarly applying Pascal, that $ N_{2}$ is also where $ BX$ intersects the circle - this means $ N_{1} = N_{2}$, which solves our problem. In the original problem (with $ BCDE$ concylic), this gives us the cute result, that $ N$ is the Miquel point of $ BCDE$.

Excuse me,how does the solution of v_Enhance works and what does this transform preserves?

v_Enhance wrote: Drop the condition that $AEDC$ is cyclic (this is the hardest part). Where does the motivation for this come from?

TAN768092100853 wrote: Excuse me,how does the solution of v_Enhance works and what does this transform preserves? When the problem statement is purely intersections, or a circle is involved in the manner it is here, we can take a projective transformation. Essentially, this let's us change our configuration into something easier to work with (think random quadrilateral ----> square) while preserving the properties that are important to what we are working with. In this manner, we can prove something about a more complicated figure, by assuming it's a simpler figure.

Proof: Let $K'$ be the intersection of $GM$ with the circumcircle. By Pascal's on $(AGK'BCC)$, we see that $D$, $M$, and the intersection of $AC$ and $K'B$ are collinear. Similarly, if $K''$ is the intersection of $LH$ with the circumcircle, $L$, $E$, and the intersection of $AC$ and $BK''$ is collinear. Since $L, E, D, M$ lie on one line, we must have $K'=K''$, so the intersection of $LH$ and $GM$ lies on the circumcircle.

bary-bash practice: let $A(1,0,0),B(0,1,0),C(0,0,1)$ then circle through $A$ and $C$ has equation: $a^2yz+b^2xz+c ^2xy=(x+y+z)vx$ thus we get: $E(\frac{v}{c^2},1-\frac{v}{c^2},0)$ and $D(0,1\frac{v}{a^2},\frac{v}{a^2})$ lines $AD$ and $CE$ have equations : $\frac{y}{z}=\frac{a^2}{v}-1$ and $\frac{y}{x}=\frac{c^2}{v}-1$ intersecting with $(ABC)$ we get : $G(1,\frac{c^2}{v}-1,\frac{(v-c^2)c^2}{a^2c^+v(b^2-a^2)}=(1,\frac{c^2}{v}-1,A)$ and $H(\frac{a^2(v-a^2}{a^2c^2+v(b^2-c^2},\frac{a^2}{v}-1,1)=(B,\frac{a^2}{v}-1,1)$ $DE$ has equation $x(v-c^2)+yv+z(v-a^2)=0$ and intersecting with the tangents ,we get : $L(\frac{b^2v+c^2(a^2-v}{c^2(v-c^2},-\frac{b^2}{c^2},1)=(C,-\frac{b^2}{c^2},1)$ and $M(1,-\frac{b^2}{a^2},\frac{a^2(v-c^2)-b^2v}{a^2(a^2-v)})=(1,-\frac{b^2}{a^2},D)$ we calculate equations of $LH$ and $MG$ and it remains to verify that the following system has a solution:which can be easily verified $$x(\frac{b^2}{a^2}D+A(\frac{c^2}{v}-1)+(A-D)y-z(\frac{c^2}{v}+\frac{b^2}{a^2}-1)=0$$$$x(\frac{a^2}{v}+\frac{b^2}{c^2}-1)+(C-B)y-z(\frac{b^2}{c^2}B+C(\frac{a^2}{v}-1)=0$$$$a^2yz+b^2xz+c^2xy$$

A solution without Pascal. By Reim's theorem, $DE \parallel GH$. Let $LH$ meet $(ABC)$ in $P$. Then $\angle APL = \frac{1}{2} \cdot \widehat{HA} = \angle AED$, so $LPEA$ is cyclic. $DE$ is antiparallel to $AC$ in triangle $ABC$, so if $P'$ is the Miquel point of $ADEC$ and $O$ is the circumcenter of $ADEC$, we have $L \in (AOE)$, so $P \in (AOE)$, so $P \equiv P'$. Thus by symmetry, we have $LH$ and $MG$ meeting on the Miquel point of $ADEC$, which is obviously on $(ABC)$.

Super-fast solution by Pascal's theorem! Let $(BDE)$ meet $\Gamma$ again at $P$. So, $BP, AC$ and $DE$ concur (at the radical center). By Pascal's theorem on $PHCAAB$, $P,H$ and $L$ are collinear. Similarly, $P,G$ and $M$ are collinear. Hence, we are done. $\square$

This feels like cheating: We will drop the $ACDE$ cyclic condition, and the problem will still be true. Let $P$ be $\overline{B(AD\cap CE)}\cap\Gamma$. Projective transform $BAPC$ to a rectangle such that $\Gamma$ goes to a circle. This can be done because there exists a rectangle whose cross ratio is anything, in particular $(BC;AP)$. Essentially, this lets us assume that $AC$ is a diameter of $\Gamma$ and $DE\parallel AC$. Solving this simpler problem will then be sufficient. [asy][asy] unitsize(0.2inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -20.537925622680746, xmax = 42.02988030821132, ymin = -27.05257022379351, ymax = 14.794727429489539; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-7.21179552830893,-7.591545392800713)--(0.6294944357574127,3.296877685917595), linewidth(2) + wrwrwr); draw((0.6294944357574127,3.296877685917595)--(14.909549344228667,-6.986893474570099), linewidth(2) + wrwrwr); draw((14.909549344228667,-6.986893474570099)--(-7.21179552830893,-7.591545392800713), linewidth(2) + wrwrwr); draw(circle((3.848876907959869,-7.289219433685406), 11.06480346540287), linewidth(2) + wrwrwr); draw((-7.404651151406892,-0.5358731613417113)--(14.716693721130705,0.06877875688890246), linewidth(2) + wrwrwr); draw((14.716693721130705,0.06877875688890246)--(14.909549344228667,-6.986893474570099), linewidth(2) + wrwrwr); draw((-7.21179552830893,-7.591545392800713)--(-7.404651151406892,-0.5358731613417113), linewidth(2) + wrwrwr); draw((14.909549344228667,-6.986893474570099)--(-4.091644383258834,0.41649256143184), linewidth(2) + wrwrwr); draw((-7.21179552830893,-7.591545392800713)--(9.541216170387914,2.1990443405470286), linewidth(2) + wrwrwr); /* dots and labels */ dot((-7.21179552830893,-7.591545392800713),dotstyle); label("$A$", (-8.390020445189366,-7.550916948477138), NE * labelscalefactor); dot((0.6294944357574127,3.296877685917595),dotstyle); label("$B$", (0.7920082173960934,3.7031621291533527), NE * labelscalefactor); dot((14.909549344228667,-6.986893474570099),dotstyle); label("$C$", (15.052592556190323,-6.575834284711319), NE * labelscalefactor); dot((5.463252138911206,-0.18414940978282468),dotstyle); label("$D$", (5.626793221146844,0.20911591732583595), NE * labelscalefactor); dot((-2.0247597677039026,-0.38882235300379997),linewidth(4pt) + dotstyle); label("$E$", (-2.539524306196861,-0.03465474861561871), NE * labelscalefactor); dot((-7.404651151406892,-0.5358731613417113),linewidth(4pt) + dotstyle); label("$L$", (-7.699336873224973,-0.07528319293919448), NE * labelscalefactor); dot((14.716693721130705,0.06877875688890246),linewidth(4pt) + dotstyle); label("$M$", (14.890078774551643,0.4122581389437148), NE * labelscalefactor); dot((9.541216170387914,2.1990443405470286),linewidth(4pt) + dotstyle); label("$G$", (9.689637762113861,2.524937243769655), NE * labelscalefactor); dot((-4.091644383258834,0.41649256143184),linewidth(4pt) + dotstyle); label("$H$", (-4.367804349632018,1.2248270254152305), NE * labelscalefactor); dot((6.485176119623775,3.4569334683801336),dotstyle); label("$B'$", (6.642504356388598,3.865675906447656), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here $B'$ is the reflection of $B$ about the perpendicular bisector of $AC$. We will show that $B'$ is the desired concurrence point. Note that $\angle EBA=\angle EMC=90^\circ$, so $EBMC$ is cyclic, so $DE\cdot DM=DB\cdot DC$. But $DB\cdot DC=DG\cdot DA$, so $DE\cdot DM=DG\cdot DA$, so $GEAM$ is cyclic. Thus, \[\angle DGM=\angle DEA=\angle CAB=\angle B'CA=\angle B'GD,\]where all angles are directed, so $B',G,M$ collinear. Similarly, $B',H,L$ collinear, so $GM$ and $HL$ concur on $\Gamma$.

A simple angle-chasing solution: Let $\odot (BDE) \cap \Gamma = X$. Note that $\angle HAB = \angle HCB =\angle ECD = \angle EAD = \angle BAG \Rightarrow \overarc{HB}=\overarc{BG}$ Also, $\angle BXA = \angle BCA = \angle BED =180^{\circ}-\angle BXD \Rightarrow BX$ is the external angle bisector of $\angle AXD$. This means that $XA$ and $XD$ are isogonal w.r.t. $\angle GXH$. Now, Let $GM \cap \Gamma = X'$. Notice that $\angle MDC = \angle EDB =\angle BAC=\angle MCD \Rightarrow MC=MD$. But, $MG \cdot MX'=MC^2=MD^2 \Rightarrow MD$ is tangent to $\odot (GDX')$ $\Rightarrow \angle GX'D = \angle MDG = \angle ADE = \angle ACH = \angle AXH =\angle GXD \Rightarrow G,X,X',D$ are concyclic $\Rightarrow D$ lies on $\Gamma$, which is not true. Thus, We get that $X'=X$ $\Rightarrow M,G,X$ are collinear. Similarly, $L,H,X$ are collinear. Hence, $LH$ and $MG$ meet on $\Gamma$.

Rename the problem to be $A$-indexed, rename $G$ and $H$ to $P$ and $Q$, and rename $L$ and $M$ to $R$ and $S$, and let $D$ and $E$ be any points on $\overline{AB}$ and $\overline{AC}$. Notice that $X=\overline{BC} \cap \overline{DE}$ is outside of $\Gamma$, so there exists a homography fixing $\Gamma$ and taking $X$ to a point at infinity (by a well-known lemma), which means $\overline{BC} \parallel \overline{DE}$. Then, we can angle chase to show that $BEQR$ and $CDPS$ are cyclic, and angle chase to show that both $\overline{QR}$ and $\overline{PS}$ pass through the reflection of $A$ over the perpendicular bisector of $\overline{BC}$.

Apparently solved this in the past but because of my bad memory, here's a different solution than last time. Let $X = MG \cap \Gamma \neq G$. Note that since $MD^2 = MC^2 = MG.MX$, we have $MD$ is tangent to $(GDX)$. To finish, $\angle BXD = \angle BXG + \angle GXD = \angle EAD + \angle GDM = \angle EAD + \angle ADE = \angle BED$, so $X$ lies on $(BDE)$, similarly $LH \cap \Gamma$ does too, so they're the same point. $\blacksquare$

i heckin love geometry After being told by a certain "Even-chan" to ignore the fact that $AEDC$ is cyclic, we employ a projective transformation which fixes $\Gamma$ and sends $F$ to the center of $\Gamma$, where $F$ is the point on $\overline{AC}$ such that $\overline{AD},\overline{CE},\overline{BF}$ concur (I hate $B$-indexed geo). Then $\triangle ABC$ is right with $\angle B=90^\circ$, and it is well-known (by Ceva's) that $\overline{DE} \parallel \overline{AC}$. Then $\angle ALD=\angle AHE=\angle ABD=90^\circ$ hence $ALHE$ and $ALBD$ are cyclic. Likewise, $CMGD$ and $CMBE$ are cyclic. Thus, we have $$\angle BHL=\angle BHE=\angle BCE=\angle BCH=\angle BAH=\angle EAH=\angle ELH=\angle MLH,$$and likewise $\angle BLH=\angle LMG$. Thus if $\overline{LH} \cap \overline{MG}=T$, then $\triangle BLM$ and $\triangle TML$ are simply reflections of each other over the perpendicular bisector of $\overline{LM}$, hence $T$ lies on $\Gamma$ as well. $\blacksquare$

can we prove this with pappus's hexagon theorem?

Let $F_1 = \overline{LH} \cap \Gamma$ and $F_2 = \overline{GF} \cap \Gamma$. By Pascal on $F_1HCAAB$, $\overline{BF_1} \cap \overline{CA}$ lies on $\overline{EL}$. By similar logic, $\overline{BF_2} \cap \overline{CA}$ lies on $\overline{EM} = \overline{EL}$, hence $F_1=F_2$.

Let $T_1=\overline{LH} \cap \Gamma$ and $T_2=\overline{GM} \cap \Gamma$. By Pascal's theorem on $T_1HCAAB$, $\overline{T_1B}$, $\overline{AC}$, and $\overline{DE}$ concur. Similarly, Pascal on $T_2GACCB$ implies that $\overline{T_2B}$, $\overline{AC}$, and $\overline{DE}$ concur. We perform the following casework. If $T_1=T_2$, then we are done by phantom point. If $T_1 \neq T_2$, then it follows that $\overline{AC} \cap \overline{DE}=B$, but this is impossible since $\triangle ABC$ is acute. We are done.

huh? Let $T = \overline{AC} \cap \overline{DE}$, $X_1 = \overline{LH} \cap \Gamma, X_2 = \overline{MG} \cap \Gamma$. By Pascal's on $AABX_1HC$ and $CCBX_2GA$ we find that $B$ lies on both $TX_1$ and $TX_2$, which is enough to imply $X_1 = X_2$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -6.6496769233372595, xmax = 12.941815109060572, ymin = -3.613273683652939, ymax = 6.384744374178368; /* image dimensions */ draw((1.0741007882469018,5.864779964745696)--(-1.569203339209051,-0.20761001762715192)--(7.030796660790949,-0.3076100176271519)--cycle, linewidth(0.4)); /* draw figures */ draw((1.0741007882469018,5.864779964745696)--(-1.569203339209051,-0.20761001762715192), linewidth(0.4)); draw((-1.569203339209051,-0.20761001762715192)--(7.030796660790949,-0.3076100176271519), linewidth(0.4)); draw((7.030796660790949,-0.3076100176271519)--(1.0741007882469018,5.864779964745696), linewidth(0.4)); draw(circle((2.751502595822466,1.5231003950833546), 4.65444500554912), linewidth(0.4)); draw((xmin, -0.07090363035037757*xmin + 3.1275870612972865)--(xmax, -0.07090363035037757*xmax + 3.1275870612972865), linewidth(0.4)); /* line */ draw((xmin, -2.4964927138011253*xmin-4.1251147204349445)--(xmax, -2.4964927138011253*xmax-4.1251147204349445), linewidth(0.4)); /* line */ draw((xmin, 2.3375046294911606*xmin-16.74212976123699)--(xmax, 2.3375046294911606*xmax-16.74212976123699), linewidth(0.4)); /* line */ draw((xmin, -0.48193519025145726*xmin + 3.0807783087104443)--(xmax, -0.48193519025145726*xmax + 3.0807783087104443), linewidth(0.4)); /* line */ draw((xmin, 0.5492122081505503*xmin + 0.6542156133370679)--(xmax, 0.5492122081505503*xmax + 0.6542156133370679), linewidth(0.4)); /* line */ draw((xmin, 0.23740182513473312*xmin + 4.049444576888316)--(xmax, 0.23740182513473312*xmax + 4.049444576888316), linewidth(0.4)); /* line */ draw((5.420494900154072,5.336279959318405)--(8.250144775637668,2.5426218457883745), linewidth(0.4)); /* dots and labels */ dot((1.0741007882469018,5.864779964745696),dotstyle); label("$B$", (1.1377966795111414,6.009096296861006), NE * labelscalefactor); dot((-1.569203339209051,-0.20761001762715192),dotstyle); label("$A$", (-1.5061878646841411,-0.059064952111752174), NE * labelscalefactor); dot((7.030796660790949,-0.3076100176271519),dotstyle); label("$C$", (7.090373904693963,-0.1602009729279648), NE * labelscalefactor); dot((-0.11388116425441086,3.1356616492714515),dotstyle); label("$E$", (-0.061387567309669766,3.278423734823265), NE * labelscalefactor); dot((3.9885635786038995,2.844783423690976),dotstyle); label("$D$", (4.041845277233829,2.9894636753483717), NE * labelscalefactor); dot((-2.9900785055538854,3.339594482373688),linewidth(4pt) + dotstyle); label("$L$", (-2.936540159084868,3.451799770508201), NE * labelscalefactor); dot((8.250144775637668,2.5426218457883745),linewidth(4pt) + dotstyle); label("$M$", (8.304006154488519,2.6571596069522445), NE * labelscalefactor); dot((-1.3466097913198913,3.7297569546846714),linewidth(4pt) + dotstyle); label("$H$", (-1.2894678200779703,3.8418958507993066), NE * labelscalefactor); dot((6.530205590911957,4.240684245598893),linewidth(4pt) + dotstyle); label("$G$", (6.584693800612898,4.362023957854115), NE * labelscalefactor); dot((5.420494900154072,5.336279959318405),linewidth(4pt) + dotstyle); label("$K$", (5.472197571634555,5.445624180884964), NE * labelscalefactor); dot((2.3532646245666484,1.9466572741578922),linewidth(4pt) + dotstyle); label("$I$", (2.409220941200676,2.0647914850287132), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $K' = LH \cap (ABC)$, $M' = KG \cap CC$. By Pascal's on $AABKHC$, we get $BK \cap AC-L-E$, and pascal on $CCBKGA$ we get $BK \cap AC, E, M'$ and we're done.

Let $F = (BED) \cap \Gamma$. We make the following claim: Claim: $\triangle LAE$ and $\triangle MDC$ are isosceles. Proof: \[\angle LAE = \angle ACB = \angle C; \phantom{c} \angle LEA = \angle BED = \angle C, \]and a likewise similar proof ensues for $\triangle MDC$. $\square$ Now as $LA = LE$, it would suffice to show $L$ bears the same power with respect to $\Gamma$ as it does to $(HEF)$, as then $L, H, F$ would be collinear due to the existing radical axis. Claim: $LE$ is tangent to $(HEF)$. Proof: \begin{align*} \angle HFE &= \angle AFH - \angle AFE \\ &= \angle ACH - (\angle AFB - \angle EFB) \\ &= \angle ACH + \angle A - \angle C \\ &= \angle C - \angle HCB + \angle A - \angle C \\ &= \angle A - \angle DAE \\ &= \angle A - (\angle A - \angle DEC) \\ &= \angle HEL, \end{align*}as desired. $\square$ It therefore similarly follows that $M, G, F$ are collinear and hence both lines $LH$ and $MG$ meet at $F$ on $\Gamma$. $\blacksquare$

Let $LH \cap \Gamma = X$ and let $BX \cap AC = Z.$ Pascal on $AACHXB$ gives that $X,L,E$ are collinear, whence $BX$ passes through $DE \cap AC.$ Similarly, if $MG \cap \Gamma = X'$ then $BX'$ also passes through $DE \cap AC,$ so indeed $X = X'$ and we are done.

Solved with cursed_tangent1434. But we had different solutions, I had a Pascal while he had a Spiral Sim solution. This is mine. Let $T_1=\overline{LH} \cap (ABC)$ and let $T_2=\overline{GM} \cap (ABC)$. Now apply Pascal's Theorem on hexagons $BAACHT_1$, $BCCAGT_2$, to get that both $\overline{T_1B} \cap \overline{AC}$ and $\overline{T_2B} \cap \overline{AC}$ lie on $\overline{LEDM}$, which means $T_1=T_2$, as desired.

Solved with ihategeo_1969 and this was basically a who can solve first contest and not a group solve. We ended up finiding totally different solutions. Something different from Pascal Spam, I'm fairly sure this solution works. Let $T$ be the Miquel Point of $AEDC$. We start off with the following observations. Claim : Quadrilaterals $LADT$ and $MCET$ are cyclic. Proof : Simply note that, \[\measuredangle TAL = \measuredangle TCA = \measuredangle TDE = \measuredangle TDL \]so quadrilateral $LADT$ must be cyclic as claimed. The proof of the other is entirely similar. Now we are just done since, \[\measuredangle LTA = \measuredangle LDA = \measuredangle EDA = \measuredangle ECA = \measuredangle HCA = \measuredangle HTA\]from which it is clear that points $L$ , $H$ and $T$ lie on the same line. We can similarly show that poitns $M$ , $G$ and $T$ are also collinear, which implies that the lines $LH$ and $MG$ meet at $\Gamma$ as desired.

By Evan's conjecture on homography, we notice that since the cyclic condition is the only euclidean one in the problem, it may be dropped. Now the problem can be killed by projective techniques, the easiest of which is taking a homography.

Let $S$ be second intersection of $(BDE)$ with $(ABC)$. We have $$\angle{ESH} = \angle{BSH} - \angle{BSE} = 180^{\circ} - \angle{BCE} - \angle{BSE} = \angle{BDE} - \angle{BCE} = \angle{DEC} = \angle{LEH}$$Then $LE$ tangents $(SEH)$ at $E$. But $$\angle{LEA} = \angle{DEB} = \angle{ACB} = \angle{LAE}$$so $\triangle LEA$ is $L$ - isosceles. Hence $LE = LA$ or $L$ lies on radical axis of $(ABC)$ and $(SEH)$. This means $L, S, H$ are collinear. Similarly, $M, S, G$ are collinear

Here's my projective solution (pascal): Let $T = DE \cap AC$ and define $X = BT \cap (ABC)$ (with $X \neq B$). We will prove that $X, G, M$ are collinear. Notice that, by Pascal Theorem, $BXGACC$, $T = BX \cap AC, M' = XG \cap CC, D = GA \cap CB$ are collinear. Note that, $M = DT \cap CC$ which implies $M'=M$ and therefore $M$ lies on $XG$. Similarly, $L$ lies on $XH$ and we are done.