[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 = -44.66923525329801, xmax = 93.36585687586805, ymin = -30.428096417855166, ymax = 52.104634907416504; /* image dimensions */ /* draw figures */ draw(circle((-11.218083812614761,2.9417594483668497), 11.3016000483859), linewidth(0.4) + blue); draw((-12.705985161230227,14.144987252462617)--(-21.431509991591483,-1.8968446210181515), linewidth(0.4)); draw((-21.431509991591483,-1.8968446210181515)--(-6.794597208293725,-7.458189105114964), linewidth(0.4)); draw((-6.794597208293725,-7.458189105114964)--(-2.616915687561999,10.272930605709483), linewidth(0.4)); draw((-2.616915687561999,10.272930605709483)--(-12.705985161230227,14.144987252462617), linewidth(0.4)); draw((-12.705985161230227,14.144987252462617)--(-6.794597208293725,-7.458189105114964), linewidth(0.4)); draw((-2.616915687561999,10.272930605709483)--(-21.431509991591483,-1.8968446210181515), linewidth(0.4)); draw((-21.436649770682028,7.769499438704459)--(-7.661450424396113,12.20895892908605), linewidth(0.4)); draw((-21.436649770682028,7.769499438704459)--(-14.113053599942603,-4.677516863066558), linewidth(0.4)); draw((-14.113053599942603,-4.677516863066558)--(-8.419410721936497,-14.354301883266434), linewidth(0.4)); draw((-8.419410721936497,-14.354301883266434)--(-6.794597208293725,-7.458189105114964), linewidth(0.4)); draw((-2.616915687561999,10.272930605709483)--(-1.7087555017968137,14.12738836906399), linewidth(0.4)); draw((-1.7087555017968137,14.12738836906399)--(-7.661450424396113,12.20895892908605), linewidth(0.4)); draw((-21.436649770682028,7.769499438704459)--(-10.288607818960328,5.310678040541101), linewidth(0.4)); draw((-18.701885228455033,3.1215562636552585)--(-1.7087555017968137,14.12738836906399), linewidth(0.4)); draw((-15.055107931054582,9.826137297149662)--(-8.419410721936497,-14.354301883266434), linewidth(0.4)); draw(circle((-4.476906619937551,-0.2716659505488135), 14.624088806067475), linewidth(0.4) + linetype("4 4")); draw((-12.705985161230227,14.144987252462617)--(6.702723071990119,49.827796070493584), linewidth(0.4)); draw((6.702723071990119,49.827796070493584)--(-1.7087555017968137,14.12738836906399), linewidth(0.4)); draw((-8.419410721936497,-14.354301883266434)--(-1.8544530261285224,-25.5119536099739), linewidth(0.4)); draw((-1.7087555017968137,14.12738836906399)--(8.357687436321019,17.37159300905428), linewidth(0.4)); draw(circle((4.159320382991789,-4.28633761676648), 22.06110253154509), linewidth(0.4) + linetype("2 2") + red); draw((-2.616915687561999,10.272930605709483)--(8.357687436321019,17.37159300905428), linewidth(0.4)); draw((8.357687436321019,17.37159300905428)--(-1.8544530261285224,-25.5119536099739), linewidth(0.4)); /* dots and labels */ dot((-12.705985161230227,14.144987252462617),dotstyle); label("$A$", (-12.32289186271667,15.073027293129758), NE * labelscalefactor); dot((-21.431509991591483,-1.8968446210181515),dotstyle); label("$B$", (-21.06271166462862,-0.9649925495540369), NE * labelscalefactor); dot((-6.794597208293725,-7.458189105114964),dotstyle); label("$C$", (-6.46631158308495,-6.551269123971988), NE * labelscalefactor); dot((-2.616915687561999,10.272930605709483),dotstyle); label("$D$", (-2.231553534735861,11.198674185065697), NE * labelscalefactor); dot((-10.288607818960328,5.310678040541101),linewidth(4pt) + dotstyle); label("$O$", (-9.890158515792725,6.062903786004031), NE * labelscalefactor); dot((-7.661450424396113,12.20895892908605),linewidth(4pt) + dotstyle); label("$M$", (-7.277222698726265,12.910597651419584), NE * labelscalefactor); dot((-14.113053599942603,-4.677516863066558),linewidth(4pt) + dotstyle); label("$N$", (-13.764511623856786,-3.9383333069055273), NE * labelscalefactor); dot((-21.436649770682028,7.769499438704459),dotstyle); label("$S$", (-22.05382525041245,9.306548248569293), NE * labelscalefactor); dot((-18.701885228455033,3.1215562636552585),linewidth(4pt) + dotstyle); label("$E$", (-18.359674612490902,3.810372909222599), NE * labelscalefactor); dot((-15.055107931054582,9.826137297149662),linewidth(4pt) + dotstyle); label("$F$", (-14.665523974569357,10.567965539566895), NE * labelscalefactor); dot((-1.7087555017968137,14.12738836906399),linewidth(4pt) + dotstyle); label("$G$", (-1.3305411840232886,14.892824822987244), NE * labelscalefactor); dot((-8.419410721936497,-14.354301883266434),linewidth(4pt) + dotstyle); label("$H$", (-8.08813381436758,-13.669266694601314), NE * labelscalefactor); dot((-14.043215410930104,6.13880727456115),linewidth(4pt) + dotstyle); label("$T$", (-13.674410388785528,6.873814901645346), NE * labelscalefactor); dot((6.702723071990119,49.827796070493584),linewidth(4pt) + dotstyle); label("$Y$", (7.048873677603633,49.94220526570633), NE * labelscalefactor); dot((-11.81034193689303,10.871857712856887),linewidth(4pt) + dotstyle); label("$F'$", (-11.421879512004098,11.559079125350724), NE * labelscalefactor); dot((8.357687436321019,17.37159300905428),linewidth(4pt) + dotstyle); label("$G'$", (8.76079714395752,18.136469285552508), NE * labelscalefactor); dot((-17.31559008853936,0.7654407975216094),linewidth(4pt) + dotstyle); label("$E'$", (-16.91805485135079,1.4677407973699097), NE * labelscalefactor); dot((-1.8544530261285224,-25.5119536099739),linewidth(4pt) + dotstyle); label("$H'$", (-1.510743654165803,-24.75171860836596), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice geo Let $\alpha = \angle BAD = \angle NCH$, $\beta = \angle ABC = \angle MDG$ and let $x = \angle SMA = \angle SNB = \angle HNC = \angle DMG$. Easily we notice that $AFM \sim CHN$ and $BNE \sim DMG$, and from here we have that $\angle AFM = \angle NHC = \angle EHG$, which implies that $EHGF$ is cyclic. Now let $Y$ be the intersection of $AB$ and $CD$. We claim that $AC \parallel FH$. Proof of claim Notice that $\frac{AF}{AY} = \frac{\sin x \sin \alpha - \beta}{2\sin \alpha + x \sin \beta} = k$, this we get from applying LoS on $AFM$ and $ADG$. Similarly we have that $\frac{CH}{YC}=k$, again from applying LoS, but this time on $NCH$ and $YBC$. Now notice that: $$\frac{YA}{YC}=\frac{(k+1)YA}{(k+1)YC}=\frac{YA+AF}{YC+CH}=\frac{YF}{YH}$$by Thales we get the desired claim. $\blacksquare$ Similarly we have that $BD \parallel EG$. Now let $F',G',E',H'$, be the intersection of $AC$ with $FG$, $BD$ with $FG$, $BD$ with $EH$ and $AC$ with $FH$, respectively. Notice that $\angle F'G'E' = \angle FG'B = \angle FGE = \angle FHE = \angle F'H'E = \angle E'H'F'$, which implies that $F'E'H'G'$ is cyclic. Since we have that $\overleftrightarrow{F'H'}=\overleftrightarrow{AC} \parallel \overleftrightarrow{FH}$ and $\overleftrightarrow{E'G'}=\overleftrightarrow{BD} \parallel \overleftrightarrow{EG}$, we know that $FEHG$ and $FE'H'G'$ are homothethic. The center of that homothethy is $S$, just by knowing that it must be the intersection of $FF'$ and $EE'$. Thus the intersection of the diagonals of $F'E'G'H'$ must be colinear with $S$ and $T$, but notice that the intersection of the diagonals of $F'E'G'H'$ is $O$, thus $S,T,O$ are colinear points.

Can someone complete my solution, or see if it's solvable in the end at all? Take $$SN \cap AB = G \hspace{0.5cm} SN \cap CD = F$$$$SM \cap AB = R \hspace{0.5cm}SM \cap CD = E \hspace{0.5cm} SO \cap FE = K$$I used trig ceva on the triangle $SFE$ with cevians $FR$, $EG$, and $SK$ so we'll have that we need to prove that: $$\frac{sin \angle RFE \cdot sin\angle REG \cdot sin \angle OSF}{sin \angle RFS \cdot sin \angle OSE \cdot sin \angle GEF}=1$$Through angle chasing we can see that $RGFE$ is cyclic $\Rightarrow$ $\angle RFS = \angle REG$. So we need to prove that $$\frac{sin \angle RFE}{sin \angle GEF}=\frac{sin \angle OSE}{sin \angle OSF}$$We can see by the law of sines on $\triangle TEF$ and $\triangle SFE$ that $$\frac{sin \angle RFE}{sin \angle GEF}=\frac{TE}{TF} \hspace{0.5cm} \frac{sin \angle OSE}{sin \angle OSF}=\frac{KE \cdot SF}{KF \cdot SE}$$So we need to prove that $$\frac{TE}{TF} = \frac{KE \cdot SF}{KF \cdot SE}$$How can I finish the problem now?

Very nice problem. Let $AB \cap CD=E$ $AD\cap BC=F$ and $U,W$ mid points of $AB,CD$ Now $\angle SMF= \angle SNF$ $\angle UMF=\angle UNF$ $\angle WMF = \angle WNF $ then $M(F,S;U,W)=N(F,S;U,W)$ this gives $MNFSUW$ is conic.Now we will prove that $EFOMNUW$ is conic by geometric location of points $X$ $X(E,O;M,N)=-1$ So we get that $OSMNUWEF$ is conic then $S(E,O;M,N)=-1$ this gives $S-T-O$ collinear. $\blacksquare $

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