Sharygin 16 wrote: Let $ABCD$ be a cyclic quadrilateral, $E= AC \cap BD , F = AD \cap BC .$ The bisectors of angles $AFB$ and $AEB$ meet $CD$ at points $X, Y.$ Prove that $A, B, X, Y$ are concyclic. Define $K:=EX\cap AB, J:=FY\cap AB.$ We begin with the following claim about $K-E-X$ and $F-J-Y.$ Claim: $K-E-X||F-J-Y.$ Proof: This is simply angle chase. Let $\angle BDC=\theta, \angle ADB=\alpha, \angle ACB=\alpha, \angle ACD=\beta.$ Note that $\angle EDX=90-(\theta+\beta)/2, \angle DFY=90-\alpha-(\theta+\beta)/2.$ And we get that $$\angle EXD=90+\beta/2-\theta/2=\angle FYD\implies FY||EX$$ Define $G:= AB\cap CD.$ We begin with the following lemma. Lemma: $GX=GK$ Proof: Note that $$\Delta AKE\sim \Delta DXE \implies \angle GXK=\angle GKX$$ Now to prove that $AXYB$ is cyclic, by Power of Point, it's enough to show that $$GA\cdot GB=GD\cdot GC=GX\cdot GY.$$This motivates use to invert wrt $(G, GA\cdot GB).$ Let $K'$ and $X'$ be the inverted image of $K,X$ respectively. Note that after inversion, we will have $$\angle GXK=\angle GKX=\angle GX'K'\implies K'X'||KX.$$So it's enough to show that $K'-X'-F$ collinear and we will be done as by using our \claim, we get $K'=J,X'=Y.$ Define $Q$ as the Miquel point of $ABCD.$ Hence we have $QFBA$ cyclic and we also have $Q\in FG\implies GQ\cdot GF=GA\cdot GB\implies Q\text{ is the inverted image of }F.$ Now, to show that $F-K'-X',$ it is enough to show that $AGKX$ is cyclic. But first we proove the following lemma. Lemma: $AK/KB=DX/XC$ Proof: Note that $$\frac{AK}{KB}\cdot \frac{XC}{DX}=\frac{AE}{EB}\cdot \frac{EC}{ED}=\frac{AE\cdot EC}{EB\cdot ED}=1$$ Using the well know fact that there is a spiral similarity centred at $Q$ which take $DC\leftrightarrow AB\implies \Delta AQB\sim \Delta DQC.$ Using the lemma and the similarity, we get that $\angle QKG=\angle QXG.$ This implies that $$ QKXG\text { is cyclic}.$$We are done!

Very Nice Problem -- I used POP + Menelaus + Ratio Bash..

Here is a sketch of my solution. Let $AX$ and $BY$ intersect $BD$ and $AC$ at $R$ and $S$ respectively. It's easy to see that $RS || CD$ by twice Menelaus for triangle $DEC$, the well-known similarities in a cyclic quad and the angle bisectors in ratios. Then $ABSR$ is cyclic and easy angle chase reveals that $ABYX$ is also cyclic, done.

Define $f(Z)=\pm CZ/DZ$, where the sign is negative if and only if $Z$ lies on or below line $CD$. the concyclicity of $A,B,X,Y$ is equivalent to \[ f(A)f(B) = f(X)f(Y). \]By the angle bisector theorem, $f(X)=-CF/FD$ and $f(Y)=-CE/ED$. [asy][asy] size(8cm); pair A = dir(145), B = dir(85), C = dir(-20), D = dir(200), E = extension(A, C, B, D), F = extension(A, D, C, B), X = extension(F, incenter(D, F, C), C, D), Y = extension(E, incenter(E, D, C), C, D); draw(A--B--C--D--cycle, heavygreen); draw(unitcircle, deepgreen); draw(A--F--B--D^^A--C, lightolive); draw(circumcircle(A, B, X), olive+dashed); draw(E--Y^^F--X, deepgreen); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$F$", "$X$", "$Y$"}; pair[] points = {A, B, C, D, E, F, X, Y}; pair[] ll = {A, dir(70), C, D, dir(200), F, X, Y}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] It is therefore equivalent to show that \[ \frac{CF}{FD} \cdot \frac{CE}{ED} = \frac{CA}{DA} \cdot \frac{CB}{DB}. \]Let $\alpha = \angle BDC = \angle BAC$, $\beta = \angle ABD = \angle ACD$, and $\gamma = \angle ACB = \angle ADB$. Without loss of generality, assume the circumcircle of $ABCD$ has radius $1/2$. By the sine law, we have \[ CA = \sin(\alpha+\gamma), \; DA = \sin\beta, \; CB = \sin\alpha, \; DB = \sin(\beta+\gamma). \]However, by the sine law applied to $\triangle DFC$ and $\triangle DEC$ , we have \[ \frac{CF}{FD} = \frac{\sin(\alpha+\gamma)}{\sin(\beta+\gamma)}, \; \frac{CE}{ED} = \frac{\sin\alpha}{\sin\beta}. \]The desired ratio equality follows.

Denote by $M$ Miquel point of $ABCD.$ From $$\frac{|MC|}{|MD|}=\frac{|BC|}{|AD|}=\frac{|EC|}{|ED|}=\frac{|YC|}{|YD|}$$$MY$ bisects angle $CMD.$ From $$\measuredangle XYM=\measuredangle CDM+\frac{1}{2}\measuredangle DMC=\measuredangle CFM+\frac{1}{2} \measuredangle DFC=\measuredangle XFM$$$X,Y,F,M$ are concyclic. But $FM,AB,CD$ concur, so by radical axis $A,B,X,Y$ are concyclic.

Let $EY\cap AB=Y_{1}$, $FX\cap AB=X_{1}$ and $AB\cap CD=G$ (if $AB\parallel CD$, then $ABCD$ is a trapezoid, so $X\equiv Y$ and therefore $A$, $B$, $X$, $Y$ trivially lie on a circle, that is, the circumcircle of $\triangle ABX$). From the Miquel theorem for points $B\in \overline{AC}$, $C\in \overline{DG}$, $F\in \overline{AD}$ and $\triangle ADG$ we know that the circumcircles of $\triangle CBG$, $\triangle ADG$, $\triangle FDC$ and $\triangle FAB$ concur at a single point. Let $M$ be that point. $\textbf{Lemma 1}$ $M\in FG$ $\textit{Proof:}$ We know that $CBGM$, $ABCD$ and $FDCM$ are cyclic, so: $$\angle CMG=180^{\circ}-\angle CBG=\angle ABC=180^{\circ}-\angle ADC=\angle CDF=180^{\circ}-\angle CMF$$$$\Longrightarrow \angle CMG+\angle CMF=180^{\circ}\Longrightarrow M\in FG$$ $\textbf{Lemma 2}$ $\triangle MDC\sim\triangle MAB$ $\textit{Proof:}$ We use the properties of $M$, that is, that $M$ lies on the circumcircles of $\triangle CBG$, $\triangle ADG$, $\triangle FDC$ and $\triangle FAB$ and the fact that $M\in FG$ which we proved in $\textbf{Lemma 1}$: $$\angle CMD=\angle CFD=\angle BFA=\angle BMA$$$$\angle CDM=\angle CFM=\angle BFM=\angle BAM$$Since $\angle CMD=\angle BMA$ and $\angle CDM=\angle BAM$, this implies that $\triangle MDC\sim \triangle MAB$, so we've proven the lemma. Now we return back to the problem. Notice that since $ABCD$ is cyclic (by the problem statement) it follows that $\angle EDC=\angle BDC=\angle BAC=\angle EAB$ and $\angle EBA=\angle DBA=\angle DCA=\angle DCE$, so $\triangle EDC\sim\triangle EAB$, therefore $\frac{DE}{EC}=\frac{AE}{EB}$. Using the angle bisector theorem (which I've also cited, although I think it's very well-known) for $\triangle DEC$ and the angle bisector $EY$ and $\triangle AEB$ and the angle bisector $EY_{1}$ we have that: $$\frac{DY}{YC}=\frac{DE}{EC}=\frac{AE}{EB}=\frac{AY_{1}}{Y_{1}}$$From $\textbf{Lemma 2}$ we have that $\triangle MDC\sim\triangle MAB$, however we showed that $\frac{DY}{YC}=\frac{AY_{1}}{Y_{1}B}$, so $\triangle MYC\sim\triangle MY_{1}B$ and thus: $$\Longrightarrow \angle MYC=\angle MY_{1}B\Longrightarrow MGY_{1}Y\text{ is cyclic}$$ $\textbf{Lemma 3}$ $XX_{1}Y_{1}Y$ is an isosceles trapezoid. $\textit{Proof:}$ Let $\angle DAC=\angle DBC=x$, $\angle CAB=y$ and $\angle DBA=z$. Then, because $FX_{1}$ is an angle bisector in $\triangle AFB$ it follows that: $$\angle AFX_{1}=\frac{180^{\circ}-\angle FAB-\angle FBA}{2}=$$$$=\frac{180^{\circ}-(x+y)-(x+z)}{2}=90^{\circ}-x-\frac{y}{2}-\frac{z}{2}$$$$\Longrightarrow \angle FX_{1}A=180^{\circ}-\angle AFX_{1}-\angle FAX_{1}=$$$$=180^{\circ}-\left(90^{\circ}-x-\frac{y}{2}-\frac{z}{2}\right)-(x+y)=90^{\circ}-\frac{y}{2}+\frac{z}{2}$$Also because $EX_{1}$ is an angle bisector in $\triangle AEB$ we have that: $$\angle AEY_{1}=\frac{180^{\circ}-\angle EAB-\angle EBA}{2}=$$$$=\frac{180^{\circ}-y-z}{2}=90^{\circ}-\frac{y}{2}-\frac{z}{2}$$$$\Longrightarrow\angle EY_{1}A=180^{\circ}-\angle AEY_{1}-\angle EAY_{1}=$$$$=180^{\circ}-\left(90^{\circ}-\frac{y}{2}-\frac{z}{2}\right)-y=90^{\circ}-\frac{y}{2}-\frac{z}{2}$$$$\Longrightarrow \angle FX_{1}A=90^{\circ}-\frac{y}{2}-\frac{z}{2}=\angle EY_{1}A\Longrightarrow XX_{1}\parallel YY_{1}$$Also notice that: $$\angle XYY_{1}=\angle YEC+\angle YCE=\frac{1}{2}\angle CED+\angle DCA=\angle Y_{1}EB+\angle EBY_{1}=\angle YY_{1}X_{1}$$We proved that $XX_{1}\parallel YY_{1}$ and that $\angle XYY_{1}=\angle YY_{1}X_{1}\Longrightarrow XX_{1}Y_{1}Y$ is an isosceles trapezoid and more specifically, $XX_{1}Y_{1}Y$ is cyclic (obviously an isosceles trapezoid is cyclic).\newline\newline Since we showed that $GMYY_{1}$ and $XX_{1}Y_{1}Y$ are cyclic we have that: $$\angle XYM=360^{\circ}-\angle XYY_{1}-\angle MYY_{1}=(180^{\circ}-\angle XYY_{1})+(180^{\circ}-\angle MYY_{1})=$$$$=\angle XX_{1}Y_{1}+\angle MGY_{1}=180^{\circ}-\angle XFM\Longrightarrow \angle XYM+\angle XFM=180^{\circ}$$Therefore we've just shown that $XYMF$ is cyclic since $\angle XYM+\angle XFM=180^{\circ}$. Now, by power of a point we have that: $$GX\cdot GY=pow(G, (FXYM))=GM\cdot GF=pow(G, (ABMF))=GB\cdot GA$$$$\Longrightarrow GX\cdot GY=GA\cdot GB\Longrightarrow ABXY\text{ is cyclic}$$

why not length bash sometimes

Fun.

What is the motivation to take the Miquel point?

Here is my solution to the problem.

Define $G=AB\cap CD$ let line $\ell$ be the angle bisector of $G$. $\ell_E$ and $\ell_F$ be angle bisectors of $E$ and $F$. $\ell_E$ and $\ell_F$ intersects $DC$ at $Y$ and $X$ respectively. $A'$ and $B'$ be the reflections of $A$ and $B$ over $\ell$. First notice that by angle chase both $\ell_E$ and $\ell_F$ are perpendicular to $\ell$.(as well as $AA'$ and $BB'$) By DDIT on lines $AD,DB,BC,AC$ we see that $(A,B),(C,D),(E,F)$ are involution pairs. Reflecting from infinity of the line perpendicular to $\ell$ , $(D,C),(A',B'),(X,Y)$ are involution pairs. Since $GA'\cdot GB'=GA\cdot GB=GD\cdot GC$, $G$ is the center of this involution (inversion) Thus $GX\cdot GY=GA'\cdot GB'=GA\cdot GB\Rightarrow A,B,X,Y$ are concyclic. Q.E.D.

Congratulations for all,very nice...