Let $ T $ be the point such that $ TYXD_1 $ is a parallelogram. Since $ C_1D_1 $ $ \parallel $ $ CD, $ so from Reim theorem we get $ A, $ $ B, $ $ C_1, $ $ D_1 $ are concyclic. Since $ TY $ $ \stackrel{\parallel}{=} $ $ D_1X $ $ \stackrel{\parallel}{=} $ $ MC_1 $ $ \Longrightarrow $ $ TM $ $ \stackrel{\parallel}{=} $ $ YC_1 ,$ so $ \measuredangle TYD_1 $ $ = $ $ \measuredangle XD_1Y $ $ = $ $ \measuredangle YC_1X $ $ = $ $ \measuredangle TMD_1, $ hence $ D_1, $ $ M, $ $ T, $ $ Y $ are concyclic $ \Longrightarrow $ $ \measuredangle D_1YM $ $ = $ $ \measuredangle D_1TM $ $ = $ $ \measuredangle XYC_1. $ From $ \triangle YAE $ $ \sim $ $ \triangle YBX $ and $ \triangle YBF $ $ \sim $ $ \triangle YAX $ $ \Longrightarrow $ $ \tfrac{YE}{YX} $ $ = $ $ \tfrac{YA}{YB} $ $ = $ $ \tfrac{YX}{YF}, $ so we conclude that $ {YX}^2 $ $ = $ $ YE $ $ \cdot $ $ YF. $ i.e. $ XY $ is tangent to the $ \odot (EFX) $

As, $XB\cdot XD_1=\dfrac{XB\cdot XD}{2}=\dfrac{XA\cdot XC}{2}=XA\cdot XC_1$, we get $ABC_1D_1$ cyclic. Let $E^\prime, F^\prime, G^\prime$ be the midpoints of $AB,C_1D_1, YX$. It is well known that $YX$ is tangent to $(XE^\prime F^\prime)$. The midpoints of the complete quadrilateral $XC_1YD_1AB$ are collinear, whence $E^\prime-F^\prime-G^\prime$ are collinear. As $F^\prime$ and $G^\prime$ are the midpoints of $XM$ and $XY$, we infer $=F^\prime G^\prime \parallel YM$, so $EF\parallel E^\prime F^\prime$. As $\triangle{XAB}$ is similar to $\triangle{XD_1C_1}$, we get $\widehat{EXE^\prime}=\widehat{FXF^\prime}$, so $\widehat{EXF}$ and $\widehat{E^\prime X F^\prime}$ share the same bisector. But the altitudes dropped from $X$ to $EF$ and $E^\prime F^\prime$ actually coincide due to $EF\parallel E^\prime F^\prime$, hence their isogonals coincide, i.e. $(XEF)$ is tangent to $(XE^\prime F^\prime)$. Thereby, $XY$ is tangent to $(XEF)$.

My solution: Let $N$ the midpoint of $AB$, $P$ the midpoint of $C_1D_1$ and let $Q$ the midpoint of $XY$ $\Longrightarrow$ by Newton-Gauss line we get $N,P,Q$ are collinear, but $PQ\parallel MY$ $\Longrightarrow$ $NQ\parallel EF$. Since $ABC_1D_1$ is cyclic by IMO-Shortlist 2009 G4 in $ABC_1D_1$ we get $XY$ is tangent to $\odot (NXQ)$ $\Longrightarrow$ $\angle XNQ=\angle QXY$. Furthermore $\triangle XBA\sim \triangle XCD$ $\Longrightarrow$ $\angle NXA=\angle DXM$. Let $E'$ and $F'$ the intersections of $NQ$ with $XE$ and $XF$ respectively $\Longrightarrow$ $\angle FXY=\angle DXM+\angle MXY=\angle XNQ+\angle NXA$ $=\angle XE'F'=\angle XEF$ $\Longrightarrow$ $\angle XEF=\angle FXY$ hence $XY$ is tangen to $\odot (EFX)$.

My solution: Since $DD_1=D_1X$ and $CC_1=C_1X$ we get $\triangle XD_1C_1 \sim \triangle XDC$. Hence $\angle D_1C_1X= \angle DCX= \angle XBA$. Hence $D_1C_1BA$ is cyclic which implies $\angle D_1AX= \angle XBC_1$. Since $DM=MC$ and $XC_1=C_1C$ we get $\triangle CMC_1 \sim \triangle CDX$. Hence $C_1M= \frac{DX}{2} = D_1X$ and $MC_1 \parallel DB$ Since $\triangle YD_1C_1 \sim \triangle YBA$ and $\triangle D_1C_1X \sim \triangle ABX$ Hence: $\frac{YC_1}{YA} = \frac{D_1C_1}{AB} = \frac{D_1X}{XA} = \frac{MC_1}{XA}$ Hence $\triangle YXA \sim \triangle YMC_1 \sim \triangle YFB$ which implies $\angle YXE= \angle EFX$ Done!

I dislike how this problem is pretty much an instant 7 for anyone who has seen G4 2009.

Trivial angle chasing yields that $ABC_1D_1$ is cyclic.As in G2 2012 (http://www.artofproblemsolving.com/community/c6h546176p3160578) we get that $YX$ and $YM$ are isogonal with respect to $\triangle YAB $ (as $XC_1MD_1$ is a parallelogram).But $\angle EFX= \angle FYB + \angle FBY = \angle XAD_1 + \angle XYA = \angle EXY $, so YX is tangent to the circumcircle of $\triangle XEF $

Note that $C_1D_1\parallel CD$, so by Reim's Theorem $C_1D_1AB$ is cyclic. Also, $XC_1MD_1$ is a parallelogram. Other than these two facts, we don't need the points $C$ and $D$. We claim that $YM$ is isogonal to $YX$ with respect to $\angle AYB$-that is, $\angle AYM=\angle XYB$. Note that $\triangle YC_1D_1\sim \triangle YAB$ and \[ \angle D_1C_1M=\angle C_1D_1X=\angle XAB \]while $\angle C_1D_1M=\angle XBA$ similarly. So $YC_1D_1M\sim YABX$, implying $\angle AYX=\angle C_1YM$ as desired. Now we are pretty much done. Note that \[ \angle FXY=180-\angle XYD_1-\angle YD_1X=\angle BD_1A-\angle XYD_1=\angle BC_1A-\angle EYC_1=180-\angle EC_1Y-\angle EYC_1=\angle YEC_1. \](More succinctly, $BD_1$ is antiparallel to $AC_1$ and $YX$ is antiparallel to $YM$, so $\angle(YM,AC_1)=\angle(BD_1,YX)$.) Thus the circumcircle of $XEF$ is tangent to $XY$ and we're done. The isogonality lemma is the same as the one in ISL 2012 G2.

Since $C_1D_1 ||DC$ we have $\angle ABD_1 = \angle DCA = \angle D_1C_1A $, so $ABC_1D_1$ is cyclic. So $\angle C_1AD_1= \angle D_1BC_1 = \alpha $. Thus $\angle YD_1M= \angle YC_1M = \alpha $. Note that $XC_1MD_1$ is parallelogram. Translate $Y$ to $Z$ by vector $\overrightarrow{MD_1}$. Then $D_1MYZ$ is parallelogram, so $\angle D_1XZ = \angle MD_1Y = \angle D_1YZ = \alpha$ and thus $D_1XYZ$ is cyclic. So $\angle D_1YX = \angle D_1ZX = \angle MYC_1= \beta $. Now $ \angle YXF =\alpha +\beta$ and $\angle XFE = \alpha + \beta$ and we are done.

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Since $C_1D_1 \parallel CD$ we get: $\angle BD_1C_1 = \angle BDC = \angle BAC = \angle BAC_1 \implies ABC_1D_1$ is cyclic. Now it's enough to prove $\angle YXE = \angle XFE \iff \angle YAX + \angle AYX = \angle FBY + \angle FYB \iff \angle AYX = \angle FYB$. Now since $\angle XAY = \angle MC_1Y$ it's enough to prove that $\frac{YA}{YC_1} = \frac{AX}{MC_1}$. Using some power of point and sine theorem we have: $$\frac{YA}{YC_1} = \frac{YB}{YD_1} = \frac{sin \angle YD_1B}{sin \angle YBD_1} = \frac{\sin \angle XC_1B}{sin \angle XBC_1} = \frac{XB}{XC_1} = \frac{XB \cdot XD_1}{XC_1 \cdot MC_1} = \frac{AX}{MC_1}$$ Q.E.D.

Note $ABC_1D_1$ is cyclic. By content of Taiwan TST 2014, Round 2, Problem 6, applied to $\triangle YC_1D_1$, it follows that $YX$ and $YM$ are isogonal with respect to $\triangle YC_1D_1$. Then \[ \angle EXY = \angle XC_1Y + \angle C_1YX = \angle AD_1X + \angle MYA = \angle YFX \]and we're done (angles are directed). Note that the points $C$ and $D$ can be more or less deleted immediately. The "isogonality lemma" people have mentioned has appeared in other places too. In addition to the Taiwan above, it also shows itself in ELMO 2012/5 and BrMO2 2013/2, SL 2009 G4, SL 2012 G2.

Before Evan bashes it, I plan on doing the honours myself It is clear that if we let $XY$ meet $CD$ again at $P$ and $Q$ be on $CD$ such that $XP,XQ$ are isogonal in $\angle CXD$ then it suffices to show that $XQ \parallel YM$ which would indeed, yield the conclusion. Enough synthetic, we will Bary it now! Set $XDC$ as the reference triangle with $X=(1,0,0);D=(0,1,0);C=(0,0,1)$ and observe that $M=(0:1:1);C_1=(1:0:1);D_1=(1:1:0)$ Now, we see that by power of a point, we can set $A=(b^2-t:0:t)$ and $B=(c^2-t:t:0)$ and so, we have that if $Y=(p,q,r)$, then by using $A,D_1,Y$ collinear; $B,C_1,Y$ collinear, and expanding the determinants, we get that $p-q+r(1-\frac{t}{b^2})=p+q(1-\frac{1}{c^2})-r=0$ Thus, we see that $Y=(t(b^2+c^2)-b^2c^2:t(2t-b^2):t(2t-c^2))$ Now, we shift our focus to $A$ and see that since $\frac{DQ}{QC}.\frac{DP}{PC}=\frac{DX^2}{XC^2}=\frac{c^2}{b^2}$ Therefore, $Q=(0:b^2(2t-c^2):c^2(2t-b^2))$ Now, we let $XQ_{\infty}$ denote the point at infinity on the line $XQ$ in the projective plane. Clearly, by setting $XQ_{\infty}=(x+1:-x:-1)$ we get that $x=\frac{Q_y}{Q_z}=\frac{b^2(2t-c^2)}{c^2(2t-b^2)}$ Thus, we conclude that we need to show points $M=(0:1:1)$, $Y=(t(b^2+c^2)-b^2c^2:t(2t-b^2):t(2t-c^2))$ and $XQ_{\infty}=(2(t(b^2+c^2)-b^2c^2):-b^2(2t-c^2):-c^2(2t-b^2))$ For this, we evaluate the determinant, which nicely factors since we can pull the $t(b^2+c^2)-b^2c^2$ factor out and what remains to prove is that $(2t-c^2)(2t-b^2)+(2t-b^2)(-2t+c^2)=0$ This is fairly obvious and we conclude.

Notice that $$XD\times XB=XC\times XA \Longrightarrow XD_1\times XB=XC_1 \times XA$$which implies that $ABC_1D_1$ is cyclic. Then $$\angle YAX=\angle DA_1C_1=\angle D_1BC_1=\angle YBF$$Now, define a transformation $\Psi(X)$ centered at $Y$ which is a composition of homothety with ratio $\frac{YD_1}{YB}=\frac{YC_1}{YA}$ and reflection about the angle bisector of $\angle AYB$ (so $\Psi(D_1)=B$ and $\Psi(C_1)=A$). Quadrilateral $MC_1XD_1$ is a parallelogram, hence $$\angle MC_1D_1=\angle C_1XD_1=\angle AXB$$and $$\frac{AX}{XB}=\frac{C_1X}{D_1X}=\frac{MD_1}{C_1M}$$which together imply that $\Psi(M)=X$. Finally, \begin{align*} \angle YXE&=180^{\circ}-\angle AXY=\angle XAY+\angle XYA\\ &=\angle C_1BD_1+\angle MYB\\ &=\angle FBY+\angle FYB\\ &=180^{\circ}-\angle BFY=180^{\circ}-\angle BFE=\angle XFE \\ \end{align*}as desired.

After you note the obvious fact that $AD_1C_1B$ is cyclic and that we have to probe that $\angle YXB=\angle DMY$ you can bash the problem considering $AD_1C_1B$ circle the unit circle. I will denote with $d,c,d'$ and $c'$ the affixes of points $D', C', D$ and $C$, in this order. You will obtain: $x=\frac{ac(b+d)-bd(a+c)}{ac-bd}$ $y=\frac{ad(b+c)-bc(a+d)}{ad-bc}$ $m=\frac{d'+c'}{2}=d+c-x=\frac{ac(c-b)-bd(d-a)}{ac-bd}$ We must have $ \frac{d-m}{y-m}:\frac{y-x}{b-x}\in \Bbb {R}$ After not so nasty computations we have $\frac{d-m}{y-m}:\frac{y-x}{b-x}=\frac{d(ad-bc)^2(b-c)^2}{b(c-d)^2(c+d-a-b)(abc+abd-acd-bcd)}$, which is real.

By Reim's theorem, $ABC_1D_1$ is cyclic. Let $P$ be the midpoint of $AB$, $Q$ be the midpoint of $C_1D_1$, $R$ be the midpoint of $XY$, and $N$ be the reflection of $X$ over $P$. The Newton-Gauss line of complete quadrilateral $YAC_1XBD_1$ is $PQR$, so dilating it about $X$ by a factor of 2, we have that $Y$, $M$, and $N$ are collinear. By ISL 2009 G4, we have that the circumcircle of $XPQ$ is tangent to $XY$, so the circumcircle of $XMN$ is tangent to $XY$. Since $BN\parallel AC_1$ and $C_1M\parallel BD_1$, so triangles $BNF$ and $C_1EM$ are homothetic. The center of homothety is $Y$, so we have $\frac{YN}{YE}=\frac{YF}{YM}\implies YN\cdot YM=YF\cdot YE$. Now, by Power of a Point, $YX^2=YN\cdot YM=YF\cdot YE$, so we are done.

Reim's Theorem implies $ABC_1D_1$ is cyclic. Erase $C$ and $D$, and observe the problem is to prove that $\angle YEC_1=\angle YXD_1$. It will be sufficient to prove $\triangle YEC_1\sim \triangle YXD_1$. We know that $\angle YC_1E= \angle YDX$ since $ABC_1D_1$ is cyclic, so it suffices to show $YX$ and $YM$ are isogonal in $\angle AYB$. Observe that $\triangle D_1MC_1$ is similar to $\triangle BXA$ in this order, so the flip dilation mapping $\triangle YD_1C_1$ to $\triangle YBA$ maps the corresponding pair of points $M$ to $X$, thus $YX$ and $YM$ are isogonal as desired.

Sorry for making a mistake, this post have been discarded.

jred wrote: ... Since $ \measuredangle XD_1Y $ ≠ $ \measuredangle YC_1X $, hence $ D_1, $ $ M, $ $ T, $ $ Y $ are not concyclic, consequently the proof is incorrect. you have to review your statement which is false : $ D_1, $ $ M, $ $ T, $ $ Y $ are effectively concyclic

Since $AX \cdot CX = BX \cdot DX$, we have $AX \cdot XC_1 = BX \cdot XD_1$, so $A, B, C_1, D_1$ are cyclic. Also, we note that $C_1M \parallel DX \equiv BD$ and $D_1M \parallel CX \equiv CA$. We will show $\triangle XAY \sim \triangle FBY$, and the desired will follow. It suffices to show $\frac{XA}{AY} = \frac{BF}{BY}$. Since $\frac{BF}{BY} = \frac{C_1M}{C_1Y} = \frac{XD_1}{C_1Y}$, it suffices to show $\frac{XA}{AY} = \frac{XD_1}{C_1Y}$. We can now show $\frac{XA}{XD_1} = \frac{AY}{C_1Y}$. Now $\frac{AY}{C_1Y} = \frac{AB}{C_1D_1} = \frac{AX}{XD_1}$, as required.

First, we note $ADX\sim BCX$ and $BC_1, AD_1$ are corresponding medians so $\angle AD_1X=\angle BC_1X\implies ABC_1D_1$ cyclic. Next, note $MD_1C_1\sim XCD\sim XBA\sim XC_1D_1$ so $XC_1MD_1$ is a parallelogram. Now, we have $YD_1C_1\sim YBA$ with $M,X$ in corresponding places because $MD_1C_1\sim XBA$. Thus, $\angle C_1YM=\angle AYX$ so $M,X$ lie on isogonal lines wrt $\angle AYB$. However, we also have $\angle YD_1X=\angle YC_1X$ because $\angle AD_1B=\angle AC_1B$, so $YD_1X\sim YC_1E$ and then $\angle YEX=\angle YXF$ because $\angle YEC_1=\angle YXD_1$, so we are done.

TelvCohl wrote: Let $ T $ be the point such that $ TYXD_1 $ is a parallelogram. Since $ C_1D_1 $ $ \parallel $ $ CD, $ so from Reim theorem we get $ A, $ $ B, $ $ C_1, $ $ D_1 $ are concyclic. Since $ TY $ $ \stackrel{\parallel}{=} $ $ D_1X $ $ \stackrel{\parallel}{=} $ $ MC_1 $ $ \Longrightarrow $ $ TM $ $ \stackrel{\parallel}{=} $ $ YC_1 ,$ so $ \measuredangle TYD_1 $ $ = $ $ \measuredangle XD_1Y $ $ = $ $ \measuredangle YC_1X $ $ = $ $ \measuredangle TMD_1, $ hence $ D_1, $ $ M, $ $ T, $ $ Y $ are concyclic $ \Longrightarrow $ $ \measuredangle D_1YM $ $ = $ $ \measuredangle D_1TM $ $ = $ $ \measuredangle XYC_1. $ From $ \triangle YAE $ $ \sim $ $ \triangle YBX $ and $ \triangle YBF $ $ \sim $ $ \triangle YAX $ $ \Longrightarrow $ $ \tfrac{YE}{YX} $ $ = $ $ \tfrac{YA}{YB} $ $ = $ $ \tfrac{YX}{YF}, $ so we conclude that $ {YX}^2 $ $ = $ $ YE $ $ \cdot $ $ YF. $ i.e. $ XY $ is tangent to the $ \odot (EFX) $ Please i just want to know why you decide to take the point T to create a parallelogram In fact i would like to know what was yoyr intuition

Since $ABCD$ is cyclic and $C_1D_1 ||CD$, by Reim, $ABC_1D_1$ is cyclic too. Construct $Z$ such that $BXAZ$ is a parallelogram. By Newton-Gauss line on the complete quadrilateral $XD_1YC_1$, the three points $\{$ (Midpoint of $\overline{XM}$), (Midpoint of $\overline{XY}$), (Midpoint of $\overline{AB}$ = Midpoint of $\overline{XZ}$) $\}$ are colinear. Taking a homothety of factor $2$ at $X$ gives $Y, M, Z$ are colinear. Now by paralleogram isogonality lemma, $\angle YMY = \angle XYB$, and since $(A,B,C_1, D_1)$ is cylic, $\angle XBY = \angle XAY$. So $\angle AEY = \angle BXY \Rightarrow \angle YXE = \angle YXF$, so $\Delta YXF \sim \Delta YXE$, or $YX^2 = YE \cdot YF$, which yields the desired result.

[asy][asy] import markers; unitsize(1.7inch); pair A, B, C, D, X, R, S, M, Y, E, F; A = dir(200); B = dir(120); C = dir(40); D = dir(340); X = extension(A, C, B, D); R = (X+C)/2; S = (X+D)/2; M = (C+D)/2; Y = extension(B, R, A, S); E = extension(Y, M, A, X); F = extension(Y, M, B, D); draw(A--B--C--D--cycle); draw(A--C, lightblue); draw(B--D, lightblue); draw(B--Y, deepmagenta); draw(A--Y, deepmagenta); draw(Y--E, lightred); draw(Y--X, lightred); draw(M--R, lightblue); draw(M--S, lightblue); draw(circumcircle(A, B, C), dashed+heavygreen); draw(anglemark(Y, A, E)); draw(anglemark(X, B, Y)); markscalefactor = .02; draw(anglemark(Y, S, M)); draw(anglemark(M, R, Y)); markangle(n=2, M, Y, S); markangle(n=2, R, Y, X); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, dir(90)); dot("$C_1$", R, dir(90)); dot("$D_1$", S, dir(270)); dot("$M$", M, dir(320)); dot("$Y$", Y, dir(D)); dot("$E$", E, dir(270)); dot("$F$", F, dir(250)); [/asy][/asy] By power of a point at $X$, we get that $ABC_1D_1$ is cyclic, so let $\alpha= \angle YAC_1 = \angle YBD_1$. Since $C_1M\parallel BD_1$ and $D_1M\parallel AC_1$, $\alpha = \angle YC_1M = \angle YD_1M$. By Law of Sines on $\triangle YMC_1$ and $\triangle YMD_1$, $\frac{\sin\alpha}{YM} = \frac{\sin\angle MYC_1}{MC_1} = \frac{\sin\angle MYD_1}{MD_1}$, so \[\frac{\sin\angle MYC_1}{\sin\angle MYD_1} = \frac{MC_1}{MD_1} = \frac{XD}{XC}= \frac{XA}{XB}.\]By Law of Sines on $\triangle YXA$ and $\triangle YXB$, $\frac{\sin\alpha}{YX} = \frac{\sin\angle XYA}{XA} = \frac{\sin\angle XYB}{XB}$, so \[\frac{\sin\angle XYA}{\sin\angle XYB} = \frac{XA}{XB}.\]Since $\frac{\sin\angle MYC_1}{\sin\angle MYD_1} = \frac{\sin\angle XYA}{\sin\angle XYB}$, $YX$ and $YM$ are isogonal with respect to $\angle BYA$. Finally, $\triangle YEA\sim\triangle YXB\implies \angle YEA = \angle YXB\implies \angle FEX =\angle YXF$ and we are done.

I'm not sure if this has been posted before, but here's my solution: The crucial claim is that $XY$ and $MY$ are isogonal w.r.t. $\angle AYB,$ but this clearly follows from the fact that $XC_1MD_1$ is a parallelogram and $C_1D_1$ and $AB$ are antiparallel w.r.t. $\angle AYB$ (the former is due to midpoints, the latter is due to Reim's Theorem). As a result, we get that $\triangle XYB\stackrel{-}{\sim}\triangle EYA\implies \measuredangle BXY=\measuredangle YEA$ and thus $\triangle YEX\stackrel{-}{\sim}\triangle YXF,$ finishing the problem. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ 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 = -12.35, xmax = 20.37, ymin = -11.8, ymax = 7.48; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-3.67,5.22)--(-5.61,-4.5)--(6.55,-2.92)--(3.6350372297187024,5.320685423787964)--cycle, linewidth(1) + zzttqq); draw((-3.67,5.22)--(2.00850753834272,0.6971965399109836)--(8.670257635954904,1.7840389594611783)--cycle, linewidth(1) + qqwuqq); draw((-5.61,-4.5)--(0.45063640820267725,1.9380058353454208)--(8.670257635954904,1.7840389594611783)--cycle, linewidth(1) + qqwuqq); draw((0.45063640820267725,1.9380058353454208)--(-1.2120699326557094,0.17177001390193963)--(8.670257635954904,1.7840389594611783)--cycle, linewidth(1)); draw((0.45063640820267725,1.9380058353454208)--(2.00850753834272,0.6971965399109836)--(8.670257635954904,1.7840389594611783)--cycle, linewidth(1)); /* draw figures */ draw(circle((0.06313593329279922,-0.5786917397724317), 6.8964577711566495), linewidth(1)); draw((-3.67,5.22)--(6.55,-2.92), linewidth(1)); draw((3.6350372297187024,5.320685423787964)--(-5.61,-4.5), linewidth(1)); draw(circle((0.4199750494925035,0.30112937423341163), 1.6371636044881186), linewidth(1)); draw((-3.67,5.22)--(-5.61,-4.5), linewidth(1) + zzttqq); draw((-5.61,-4.5)--(6.55,-2.92), linewidth(1) + zzttqq); draw((6.55,-2.92)--(3.6350372297187024,5.320685423787964), linewidth(1) + zzttqq); draw((3.6350372297187024,5.320685423787964)--(-3.67,5.22), linewidth(1) + zzttqq); draw((2.04283681896069,3.6293456295666924)--(5.092518614859351,1.200342711893982), linewidth(1)); draw((5.092518614859351,1.200342711893982)--(3.5003182041013385,-0.49099708232728956), linewidth(1)); draw(circle((-2.087194841319807,-0.14951049463370084), 5.597938470752076), linewidth(1)); draw((2.04283681896069,3.6293456295666924)--(3.5003182041013385,-0.49099708232728956), linewidth(1)); draw((-3.67,5.22)--(2.00850753834272,0.6971965399109836), linewidth(1) + qqwuqq); draw((2.00850753834272,0.6971965399109836)--(8.670257635954904,1.7840389594611783), linewidth(1) + qqwuqq); draw((8.670257635954904,1.7840389594611783)--(-3.67,5.22), linewidth(1) + qqwuqq); draw((-5.61,-4.5)--(0.45063640820267725,1.9380058353454208), linewidth(1) + qqwuqq); draw((0.45063640820267725,1.9380058353454208)--(8.670257635954904,1.7840389594611783), linewidth(1) + qqwuqq); draw((8.670257635954904,1.7840389594611783)--(-5.61,-4.5), linewidth(1) + qqwuqq); draw((0.45063640820267725,1.9380058353454208)--(-1.2120699326557094,0.17177001390193963), linewidth(1)); draw((-1.2120699326557094,0.17177001390193963)--(8.670257635954904,1.7840389594611783), linewidth(1)); draw((8.670257635954904,1.7840389594611783)--(0.45063640820267725,1.9380058353454208), linewidth(1)); draw((0.45063640820267725,1.9380058353454208)--(2.00850753834272,0.6971965399109836), linewidth(1)); draw((2.00850753834272,0.6971965399109836)--(8.670257635954904,1.7840389594611783), linewidth(1)); draw((8.670257635954904,1.7840389594611783)--(0.45063640820267725,1.9380058353454208), linewidth(1)); /* dots and labels */ dot((-3.67,5.22),dotstyle); label("$A$", (-3.59,5.42), NE * labelscalefactor); dot((-5.61,-4.5),dotstyle); label("$B$", (-5.53,-4.3), NE * labelscalefactor); dot((6.55,-2.92),dotstyle); label("$C$", (6.63,-2.72), NE * labelscalefactor); dot((3.6350372297187024,5.320685423787964),dotstyle); label("$D$", (3.71,5.52), NE * labelscalefactor); dot((0.45063640820267725,1.9380058353454208),linewidth(4pt) + dotstyle); label("$X$", (0.53,2.1), NE * labelscalefactor); dot((3.5003182041013385,-0.49099708232728956),linewidth(4pt) + dotstyle); label("$C_{1}$", (3.59,-0.34), NE * labelscalefactor); dot((2.04283681896069,3.6293456295666924),linewidth(4pt) + dotstyle); label("$D_{1}$", (2.13,3.78), NE * labelscalefactor); dot((5.092518614859351,1.200342711893982),linewidth(4pt) + dotstyle); label("$M$", (5.17,1.36), NE * labelscalefactor); dot((8.670257635954904,1.7840389594611783),linewidth(4pt) + dotstyle); label("$Y$", (8.75,1.94), NE * labelscalefactor); dot((2.00850753834272,0.6971965399109836),linewidth(4pt) + dotstyle); label("$E$", (2.09,0.86), NE * labelscalefactor); dot((-1.2120699326557094,0.17177001390193963),linewidth(4pt) + dotstyle); label("$F$", (-1.13,0.34), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

One line proof : Trivially, $\triangle D_1MC_1\sim \triangle BXA$ and hence, so the flip dilation takes $M\mapsto X\implies YX$ and $YM$ are isogonal $\implies \angle AFM +\angle YAE=\angle BYX +\angle YBX\implies \angle YXF=\angle FEX$ completing the proof.

Note $M$ is such that $MD_1XC_1$ is a parallelogram, so by the first isogonality lemma, $YM$ and $YX$ are isogonal. Thus by reflection over the angle bisector of $\angle AYB$ and the observation that PoP yields $ABC_1D_1$ cyclic, we have the angle equality that implies the result.

First we claim that $A,B,D_1,C_1$ are cyclic. Because $$AX\cdot XC=BX\cdot XD\Longrightarrow AX\cdot XC_1=BX\cdot XD_1$$It is sufficient to prove that $AX, AM$ are isogonal wrt $\angle D_1YC_1$. Since $$\measuredangle MC_1B=\measuredangle MC_1X+\measuredangle AC_1B=\measuredangle XD_1M+\measuredangle AD_1B=\measuredangle AD_1M$$Therefore we have $\measuredangle YC_1M=\measuredangle MD_1Y$. And $D_1XC_1M$ is a parallelogram (This is well known lemma from ABJTOG). Thereofore, [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); size(6cm); 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 = 0., xmax = 8., ymin = -2., ymax = 6.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((1.903796921904145,-0.0897114233113755),0.3598399232100768,46.56657683511538,83.69856595615217)--(1.903796921904145,-0.0897114233113755)--cycle, linewidth(0.4) + qqwuqq); draw(arc((6.526069133730269,-0.09360208145168347),0.3598399232100768,125.8069377977146,162.93892691875138)--(6.526069133730269,-0.09360208145168347)--cycle, linewidth(0.4) + qqwuqq); draw(arc((2.517033766066459,5.463644084718064),0.3598399232100768,-133.43342316488463,-96.30143404384783)--(2.517033766066459,5.463644084718064)--cycle, linewidth(0.4) + qqwuqq); draw(arc((2.517033766066459,5.463644084718064),0.3598399232100768,-54.19306220228541,-17.061073081248626)--(2.517033766066459,5.463644084718064)--cycle, linewidth(0.4) + qqwuqq); /* draw figures */ draw((5.488261352195455,-1.1897714475160954)--(2.517033766066459,5.463644084718064), linewidth(0.8) + ffxfqq); draw((2.517033766066459,5.463644084718064)--(2.9416047034389594,1.0064579427530365), linewidth(0.8) + ffxfqq); draw((2.517033766066459,5.463644084718064)--(6.526069133730269,-0.09360208145168347), linewidth(0.8)); draw((2.517033766066459,5.463644084718064)--(1.903796921904145,-0.0897114233113755), linewidth(0.8)); draw((2.517033766066459,5.463644084718064)--(xmin, 1.0562354470336373*xmin + 2.805063799618098), linewidth(0.8)); /* ray */ draw((2.517033766066459,5.463644084718064)--(xmax, -0.3068966216845171*xmax + 6.236113244189718), linewidth(0.8)); /* ray */ draw((-2.4685056683225555,2.6668025387870116)--(13.000822169198551,-2.080681914278472), linewidth(0.8)); draw((1.548655022963982,-5.3509232998084455)--(11.580480469827869,5.245046335623245), linewidth(0.8)); draw((-3.0422339019648357,1.428208727281503)--(11.90231577179704,-3.1582230801924682), linewidth(0.8)); draw((-2.2432332900649916,-4.4699517331125955)--(9.17268766811453,7.587948643450818), linewidth(0.8)); label("$P\infty$",(1.3540627498318878,-1.1043694259010046),SE*labelscalefactor); label("$Q\infty$",(7.267432154584149,-1.1643427464360174),SE*labelscalefactor); /* dots and labels */ dot((1.293002033235213,-5.620953049700098),dotstyle); label("$A$", (1.3540627498318878,-3.1794463164124487), NE * labelscalefactor); dot((5.488261352195455,-1.1897714475160954),linewidth(4.pt) + dotstyle); label("$X$", (5.516211194961776,-1.6561239748231227), NE * labelscalefactor); dot((2.9416047034389594,1.0064579427530365),linewidth(4.pt) + dotstyle); label("$M$", (2.5535291605321437,1.3785260442485268), NE * labelscalefactor); dot((6.526069133730269,-0.09360208145168347),linewidth(4.pt) + dotstyle); label("$C_1$", (6.535757644056994,0.3589795951533085), NE * labelscalefactor); dot((1.903796921904145,-0.0897114233113755),linewidth(4.pt) + dotstyle); label("$D_1$", (1.2221214446548596,-0.24075361019681982), NE * labelscalefactor); dot((2.517033766066459,5.463644084718064),linewidth(4.pt) + dotstyle); label("$Y$", (2.409593191248113,5.6966051227694505), NE * labelscalefactor); dot((9.17268766811453,7.587948643450818),dotstyle); label("$L$", (9.402482365630606,5.864530420267487), NE * labelscalefactor); dot((-2.2432332900649916,-4.4699517331125955),dotstyle); label("$N$", (-1.8844965590588034,-3.1674516523054463), NE * labelscalefactor); dot((1.548655022963982,-5.3509232998084455),dotstyle); label("$O$", (1.725897337148967,-3.335376949803482), NE * labelscalefactor); dot((11.580480469827869,5.245046335623245),dotstyle); label("$P$", (10.050194227408744,5.540674489378417), NE * labelscalefactor); dot((-2.4685056683225555,2.6668025387870116),dotstyle); label("$Q$", (-1.896491223165806,2.6859444319118064), NE * labelscalefactor); dot((13.000822169198551,-2.080681914278472),dotstyle); label("$R$", (10.050194227408744,-0.984422784830979), NE * labelscalefactor); dot((11.90231577179704,-3.1582230801924682),dotstyle); label("$S$", (10.050194227408744,-2.4117878135642843), NE * labelscalefactor); dot((-3.0422339019648357,1.428208727281503),dotstyle); label("$T$", (-1.812528574416788,1.6544033187095857), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The line through $Y$ parallel to $MD_1$ meets $MD_1$ at $P_{\infty}$ and similarly defined $Q_{\infty}$. Since $YP_{\infty}$ and $YQ_{\infty}$ are isogonal then by the isogonal lemma we have $M=D_1P_{\infty}\cap C_1Q_{\infty}$ and $X=D_1Q_{\infty}\cap C_1P_{\infty}$ are isogonal and we are done.

Note that $BX.XD = AX.XC \implies BX.XD_1 = AX.XC_1 \implies ABC_1D_1$ is cyclic. $\frac{YC_1}{YA} = \frac{C_1D_1}{AB} = \frac{D_1X}{XA} = \frac{MC_1}{XA} \implies YXA$ and $YMC_1$ are similar so $\angle YXE = \angle C_1ME = \angle EFX$.

Solved with jelena_ivanchic and TheProblemIsSolved ! Reim's theorem gives $ABC_1D_1$ cyclic. As $C_1M \parallel BD$ and $D_1M \parallel AC$, we get $\angle YC_1M=\angle YD_1M$ (each being equal to $\angle C_1BD_1=\angle C_1AD_1$). Hence by first isogonality lemma, as $\angle YC_1M=\angle YD_1M$ and $X$ is reflection of $M$ over midpoint $C_1D_1$; $YM,YX$ are isogonal. Hence, $\triangle YAX \sim \triangle YBF \implies \angle XFY=\angle EFY$ and we are done!

It's enough to show that $\angle XFY = \angle YXE$. As $D_1$ and $C_1$ are midpoints, by Thales's theorem we know that $C_1D_1 \parallel CD$ and of the cyclic $ABCD$ we have, $\angle BAC_1 = \angle BAC = \angle BDC = \angle BD_1C_1 \Rightarrow ABC_1D_1$ es cíclico. Then $\angle D_1AC_1 = \angle C_1BD_1$. It's clear that $C_1M \parallel XD$ y $D_1M \parallel XC$ so $XC_1MD_1$ is a parallelogram. Let $K = C_1M \cap AY $ and $L = D_1M \cap BY$. Notice that $\angle DAC = \angle DAD_1 + \angle D_1AC_1 = \angle DBC = \angle CBC_1 + \angle C_1BD_1$ but $\angle D_1AC_1 = \angle C_1BD_1 \Rightarrow \angle DAD_1 = \angle CBC_1$ and as $\angle ADD_1 = \angle ADB = \angle ACB = \angle C_1CB$ we get that $\angle DD_1Y = \angle CC_1Y$. Hence of parallelogram $XC_1MD_1$, $\angle DD_1Y = \angle D_1KC_1 = \angle CC_1Y = \angle C_1LD_1$, so $D_1C_1LK$ is cyclic and $\angle MKY = \angle MLY = \angle XD_1Y = \angle XC_1Y$ and thus $\angle C_1KL = \angle C_1D_1L = \angle XC_1D_1$ then $XD_1YC_1$ $\sim$ $MLYK$, from where $XC_1$ is homologous of $KM$. So $\angle FYB = \angle MYL = \angle XYD_1 = \angle XYA$ then by case $AA$, $\triangle AXY \sim \triangle BFY$, thus $\angle AXY = \angle BFY$ $\Rightarrow$ $\angle XFY = \angle YXE$, as desired $\blacksquare$

G #6/100 2012 G2 Vibes Claim: $ABC_1D_1$ concyclic Proof: Note $D_1C_1//DC$ so by Reim's since $ABCD$ cyclic, $ABC_1D_1$ concyclic Claim: $\triangle{YAX} \sim \triangle{YC_1M}$ Proof: Let $\angle{CAB}=a,\angle{ABD}=b,\angle{YAC}=c$ Note $$\frac{YA}{YC_1}=\frac{\sin{(a+b+c)}}{\sin{c}}$$$$\frac{AX}{MC_1}=\frac{\sin{(180-a-b-c)}}{\sin{c}}=\frac{\sin{(a+b+c)}}{\sin{c}}=\frac{YA}{YC_1}$$ Since $$\angle{YAX}=\angle{EBY}=\angle{MC_1Y}$$$$\triangle{YAX} \sim \triangle{YC_1M}$$ Note this implies $\angle{AYF}=\angle {XYB}$, $\triangle{D_1XY} \sim \triangle{C_1FY}$ Hence, $$\angle{EXY}=\angle{EFX}$$Q.E.D

Note that $ABC_1D_1$ is cyclic by Reim's theorem. Claim. $\overline{YEF}$ and $\overline{YX}$ are isogonal in triangle $YC_1D_1$. Proof. $XC_1MD_1$ is a parallelogram by definition, so use the first isogonality lemma. $\blacksquare$ Then $\angle MD_1C = \angle C_1AD_1 = \angle C_1BD_1$, so $\triangle YD_1M \sim \triangle YBX$. It follows $\angle D_1ME = \angle D_1XM$, which implies the desired tangency.

Let $M_1$ be the midpoint of $D_1C_1$. Note that $M$ is the reflection of $X$ about $M_1$ by a homothety centered at $X$ with scale factor $2$. We will make the following claims: Claim: $ABC_1D_1$ is cyclic. Proof: By PoP: \[AX\cdot XC=BX\cdot XD,\]\[2\cdot AX\cdot XC_1=2\cdot BX\cdot XD_1,\]\[\cdot AX\cdot XC_1=\cdot BX\cdot XD_1,\]which proves concyclicity $\square$ Claim: $YX$ and $YF$ are isogonal on $\triangle YC_1D_1$. Proof: Using the isogonality lemma, we first develop our initial conditions: \[\angle YD_1X=\angle YC_1X.\]We have already shown that $M$ is the reflection of $M_1$ about $D_1C_1$. Therefore, they are isogonal $\square$ We now finish by angle-chasing: \[\angle AXY=180-\angle XAD_1-\angle XYA=180-\angle FBC_1-\angle FYB=\angle BFY,\]\[\angle YFE=\angle XFE,\]as desired $\blacksquare$

Clearly, $\overline{C_1D_1} \parallel \overline{CD}$, so Reim's Theorem gives that $\overline{C_1D_1}$ is antiparallel to $\overline{AB}$. Hence, $ABC_1D_1$ is cyclic. Claim: $\overline{YF}$ and $\overline{YX}$ are isogonal in $\triangle YC_1D_1$ Proof: Notice that $MC_1XD_1$ is easily shown to be a parallelogram, so the first isogonality lemma finishes. $\square$ Now, we finish with an angle chase: \[\angle EXY = \angle XC_1Y+\angle XYC_1 = \angle YD_1F+\angle MYA = \angle YFX. \] This implies the tangency, so we are done. $\square$

The desired tangency is equivalent to $\angle EFX = \angle EXY$. We firstly note that $\triangle AXD \sim \triangle BXC$ (due to angles in the cyclic $ABCD$) and now the corresponding medians $AD_1$ and $BC_1$ yield $\angle AD_1X = \angle BC_1X$, i.e. $AD_1C_1B$ is cyclic. Hence $\angle XAY = 180^{\circ} - \angle XAD_1 = 180^{\circ} - \angle XBC_1 = \angle XBY$ and now the desired equality is equivalent to proving $\triangle XAY \sim \triangle FBY$. To get rid of the unpleasant point $F$, notice that $BF \parallel MC_1$, as $MC_1$ is a midsegment in triangle $CDX$. Hence $\triangle FBY \sim \triangle MC_1Y$ and we reduce to showing $\triangle XAY \sim \triangle MC_1Y$, i.e. $\displaystyle \frac{AY}{C_1Y} = \frac{AX}{C_1M}$ From the cyclic quadrilateral $ABC_1D_1$ we have $\triangle ABY \sim \triangle C_1D_1Y$ and $\triangle ABX \sim \triangle D_1C_1X$, hence $\displaystyle \frac{AY}{C_1Y} = \frac{AB}{C_1D_1} = \frac{AX}{D_1X}$. It remains to note that $D_1X = C_1M$ from the parallelogram $MC_1XD_1$, due to the midsegments $MC_1$ and $MD_1$ in triangle $CDX$.

Reim's induces cyclic quadrilateral $ABC_1D_1$. The first isogonality lemma tells us $YX$ and $YM$ are isogonals, implying \[\measuredangle AYX = \measuredangle FYB, \quad \measuredangle XAY = \measuredangle YBX \implies \measuredangle EXY = \measuredangle YFX. \quad \blacksquare\]

Note $ABC_1D_1$ is cyclic by PoP at $X,$ thus the first isogonality lemma on $\triangle YC_1D_1$ says that $YM,YX$ are isogonal. Then $AC_1,BD_1$ are also antiparallel in the same angle, so $\measuredangle YXF=-\measuredangle YEX$ as desired.

Note that $ABC_1D_1$ is cyclic by Reim. Then we use the first isogonality lemma on $\triangle ABY$ where we find that because $M$ is the reflection of $X$ across the midpoint of $C_1D_1,$ we see that $\angle BYF=\angle AYX.$ By our previous conyclic condition we found we have $\angle YAX=\angle FBY.$ Therefore $\angle YFX=\angle EXY,$ so we’re done.

The condition is equivalent to: $\angle YFD_1 =\angle YXC_1\iff\cancel{\angle FD_1Y} +\angle FYD_1= \cancel{\angle XC_1Y} +\angle XYC_1\iff\boxed{\angle MYA = \angle XYB}$. We'll proceed using barycentric coordinates with respect to $\Delta YAB$, in that order. Set $D_1=(d , 1-d , 0), C_1 = (e, 0 , 1-e), AB = a, YB = b, YA = c$. (Note that $D_1A =cd, C_1B=be$) Clearly $X = \Bigg(\frac{de}{d+e-de}, \frac{(1-d)e}{d+e-de}, \frac{(1-e)d}{d+e-de}\Bigg)$. But $M = \frac{\vec{D} +\vec{C}}{2} = \vec{D_1}+\vec{C_1}-\vec{X}$, that means: $y_M = \frac{(1-d)+0-y_X}{2} = \frac{(1-d)\Big(1-\frac{e}{d+e-de}\Big)}{2} = \frac{d(1-d)(1-e)}{2(d+e-de)}$. Analogously $z_M = \frac{e(1-d)(1-e)}{2(d+e-de)}$. We know that: $\angle MYA = \angle XYB\iff \frac{y_M}{z_M} = \frac{b^2}{c^2}\cdot\frac{z_X}{y_X}\iff\frac{d}{e} = \frac{b^2}{c^2}\cdot\frac{d(1-e)}{e(1-d)}\iff c(c-cd) = b(b-be)\iff$ $\iff YB\cdot YC_1 = YA\cdot YD_1\iff D_1ABC_1\text{ is cyclic} \iff$ $\iff D_1X\cdot BX = C_1X\cdot AX\iff \frac{1}{2}\cdot DX\cdot BX = \frac{1}{2}\cdot CX\cdot AX$, which is true because $ABCD$ is cyclic. $_{\blacksquare}$