Let $O$ be the circumcenter of $(BCD).$ Easy angle chasing yields that $\angle{YXC}=180^0-\angle{ADC}$, so in order to prove the problem it's enough to show that $$\angle{YAC}=180^0-\angle{ADC}=\angle{YDA}$$which is equivalent to $\triangle{YAC}\sim\triangle{YDA}$, or $YD\cdot YC=YA^2.$ Lemma. $AD\cdot CE=AC\cdot DE$ Proof. We have $DE=\frac{AD\cdot BD}{AD+AB}$, so it remains to prove that $(AD+AB)\cdot CE=AC\cdot BD.$ Now $\angle{BAD}+2\angle{BCD}=180^0$ implies that $O$ is the midpoint of arc $\overarc{BD}$ which does not contain $A$ in $(ABD).$ Hence by Ptolemy's Theorem we have $$AO\cdot BD=DO\cdot AB+BO\cdot AD=CO(AB+AD)~~~~~(1)$$But $OE\cdot OA=OB^2=OC^2$, so $\triangle{CEO}\sim\triangle{ACO}$, meaning that $$\frac{AC}{CE}=\frac{AO}{CO}.~~~~~~(2)$$By combining $(1)$ and $(2)$ we get the desired result. $\blacksquare$ Back to our original problem, let $Y'$ be the intersection of $CD$ the tangent at $A$ at $(ACD)$. Define the function $f:\mathcal{P}\to\mathbb{R}$ by $$f(T)=TE^2-\text{pow}(T,(ACD))~\forall~ T\in\mathcal{P}.$$ Note that $f$ is linear and $f(D)=DE^2,f(C)=CE^2.$ Since $\frac{Y'D}{Y'C}=\frac{AD^2}{AC^2}$, it follows that $$f(Y')=\frac{kf(C)-f(D)}{k+1}=0$$where $k=\frac{AD^2}{AC^2}$ (note that we have used the lemma). Therefore $Y'E^2=Y'D\cdot Y'C=Y'A^2$, following that $Y\equiv Y'$. In conclusion, $YA^2=Y'A^2=Y'D\cdot Y'C=YD\cdot YC$, which is what we wanted to prove. This ends our proof.

As above, it suffices to show that $YA$ is tangent to $(ACD)$. Let $I,I_A$ be the incenter, $A$-excenter of $\triangle ABD$, which both lie on $(BCD)$. Let $M$ be the midpoint of $\overline{AE}$. Then $(A,E;I,I_A)=-1$, so $MA^2=MI\cdot MI_A$, thus $XY$ is the radical axis of $A,(BCD)$. Hence $YA^2=YC\cdot YD$.

Nice tutorial for dynamic geometry! Lsway wrote: Given quadrilateral $ABCD$ such that $\angle BAD+2 \angle BCD=180 ^ \circ .$ Let $E$ be the intersection of $BD$ and the internal bisector of $\angle BAD$. The perpendicular bisector of $AE$ intersects $CB,CD$ at $X,Y,$ respectively. Prove that $A,C,X,Y$ are concyclic. Let $I$ be the incenter of $\triangle DAB$ and $\ell$ be the perpendicular bisector of $\overline{AE}$. Move point $C$ on $\odot(BID)$ with constant velocity. Note that $X, Y$ move on line $\ell$ such that $$X \overset{B}{\mapsto} C \overset{D}{\mapsto} Y$$is a projective mapping. Let $Z$ be on $\ell$ such that $\measuredangle XAZ=\measuredangle BCD$. Since $X \overset{A}{\mapsto} Z$ is also projective, we only need to verify $Y=Z$ for $C \in \{B, D, I\}$. Case 1. $C \ne I$. WLOG, $C=B$. Then $\angle AXB=\tfrac{1}{2}(\angle B-\angle D)$ and $Y$ lies on $\overline{BD}$. Thus, $\angle AYB=2\angle AXB$ and $\overline{XY}$ bisects angle $AYB$, so we are done. Case 2. $C=I$. Note that $ABEX$ and $ADEY$ are cyclic, so $\angle AXC+\angle AYC=\angle AEB+\angle ADB=180^{\circ}$ and we are done! $\blacksquare$ Remark: The dynamic viewpoint allows us to correctly guess $\overline{YA}$ is tangent to $\odot(ACD)$ far more effectively. Hence, looking at boundary cases is helpful even if one decides not to use the degree counting method.

We let $XY$ intersect $AB,AD$ at $G,H$ respectively. Let the midpoint of $AE$ be $F$. Let $\angle BAD=2a$. Then the condition $\angle BAD+2 \angle BCD=180 ^ \circ$ implies that $\angle BCD=90^{\circ}-a$. We then notice that $$\angle GBX+\angle HDY=360^{\circ}-(\angle ABC +\angle ADC)=360^{\circ}-(360^{\circ}-(\angle BAD+\angle BCD))=90^{\circ}+a$$ But $$\angle GBX+\angle GXB=180^{\circ}-\angle XGB=180^{\circ}-\angle AGF=180^{\circ}-(90^{\circ}-a)=90^{\circ}+a=180^{\circ}-\angle AHF=180^{\circ}- \angle YHD=\angle HDY+\angle HYD$$ So $\angle GBX+\angle HDY=\angle GBX+\angle GXB=\angle HDY+\angle HYD$ implying that $\angle GBX=\angle HYD$ and $\angle GXB=\angle HDY$ so $\triangle GXB\sim \triangle HDY$. This gives $GX\cdot HY=GB\cdot HD$. On the other hand, $GE\parallel AD$ and $HE\parallel AB$ so $\triangle GEB\sim \triangle HDE\sim \triangle ADB$. This gives $HD\cdot GB=GE\cdot HE$. Combining, $GX\cdot HY=GE\cdot HE$ so $\frac{GX}{GE}=\frac{HE}{HY}$.Combining with the fact that $\angle XGE=\angle EHY=90^{\circ}+a$, we have $\triangle GXE\sim \triangle HEY$. Using all the above, taking into account that $A$ and $E$ are reflections of each other in $XY$, $$\angle XAY=\angle XEY=\angle XEG+\angle HEY+\angle GEH=\angle XEG+\angle GXE+2a=180^{\circ}-\angle XGE+2a=180^{\circ}-(90^{\circ}+a)+2a=90^{\circ}+a=180^{\circ}-\angle XCY$$ Hence done.

Very nice solution jlammy.

huricane wrote: Let $O$ be the circumcenter of $(BCD).$ Easy angle chasing yields that $\angle{YXC}=180^0-\angle{ADC}$, so in order to prove the problem it's enough to show that $$\angle{YAC}=180^0-\angle{ADC}=\angle{YDA}$$which is equivalent to $\triangle{YAC}\sim\triangle{YDA}$, or $YD\cdot YC=YA^2.$ Lemma. $AD\cdot CE=AC\cdot DE$ Proof. We have $DE=\frac{AD\cdot BD}{AD+AB}$, so it remains to prove that $(AD+AB)\cdot CE=AC\cdot BD.$ Now $\angle{BAD}+2\angle{BCD}=180^0$ implies that $O$ is the midpoint of arc $\overarc{BD}$ which does not contain $A$ in $(ABD).$ Hence by Ptolemy's Theorem we have $$AO\cdot BD=DO\cdot AB+BO\cdot AD=CO(AB+AD)~~~~~(1)$$But $OE\cdot OA=OB^2=OC^2$, so $\triangle{CEO}\sim\triangle{ACO}$, meaning that $$\frac{AC}{CE}=\frac{AO}{CO}.~~~~~~(2)$$By combining $(1)$ and $(2)$ we get the desired result. $\blacksquare$ Back to our original problem, let $Y'$ be the intersection of $CD$ the tangent at $A$ at $(ACD)$. Define the function $f:\mathcal{P}\to\mathbb{R}$ by $$f(T)=TE^2-\text{pow}(T,(ACD))~\forall~ T\in\mathcal{P}.$$ Note that $f$ is linear and $f(D)=DE^2,f(C)=CE^2.$ Since $\frac{Y'D}{Y'C}=\frac{AD^2}{AC^2}$, it follows that $$f(Y')=\frac{kf(C)-f(D)}{k+1}=0$$where $k=\frac{AD^2}{AC^2}$ (note that we have used the lemma). Therefore $Y'E^2=Y'D\cdot Y'C=Y'A^2$, following that $Y\equiv Y'$. In conclusion, $YA^2=Y'A^2=Y'D\cdot Y'C=YD\cdot YC$, which is what we wanted to prove. This ends our proof. How did you get that you should use the center of BCD?

@above Given the condition $\angle BAD+2 \angle BCD=180 ^ \circ $ it's quite natural to consider the center of $(BCD)$ since then points $A,B,D,O$ are concyclic, which is nice and can be helpful.

Oh... I got it

[asy][asy] unitsize(1.5inches); pair B=dir(155); pair D=dir(25); pair C=dir(255); pair E=0.45*D+0.55*B; pair CC=2*foot(0,E,C)-C; pair A=1/conj(E); pair X=extension(C,B,E,CC+E-D); pair Y=extension(C,D,E,CC+E-B); pair XX=2*foot(circumcenter(CC,E,D),X,E)-E; pair YY=2*foot(circumcenter(CC,E,B),Y,E)-E; draw(circumcircle(B,C,D)); draw(C--B--D--cycle); draw(A--0*dir(0)); draw(X--Y); draw(X--B); draw(Y--D); draw(circumcircle(A,B,D)); draw(circumcircle(B,CC,E)); draw(circumcircle(D,CC,E)); draw(X--XX); draw(Y--YY); draw(circumcircle(YY,E,XX)); dot("$B$",B,dir(190)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-10)); dot("$E$",E,2*dir(252)); dot("$C'$",CC,2*dir(76)); dot("$A$",A,dir(A)); dot("$L$",0,dir(-90)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$X'$",XX,dir(70)); dot("$Y'$",YY,dir(YY)); [/asy][/asy] Note that $2(\pi-\angle BID)+\angle BAD=\pi$ where $I$ is the incenter of $BAD$, so the condition of the problem is equivalent to $C$ lying on the circle $(BID)$. Letting $L$ be the arc midpoint of $BD$, we have that $C$ is on the circle with center $L$ and passing through $B$ and $C$. Let $\phi$ be inversion at $E$ with power $EB\cdot ED$. Let $C'=\phi(C)$, $X'=\phi(X)$, and $Y'=\phi(Y)$. Note that $\phi(A)=L$ as $EA\cdot EL=EB\cdot ED$, so the perpendicular bisector of $AE$ gets sent to the circle with center $L$ passing through $E$. Therefore, $X'$ is the point on $(C'ED)$ other that $E$ such that $LX'=LE$, so we have that $C'EX'D$ is an isosceles trapezoid. Thus, $EX'\parallel C'D$, and we similarly derive $EY'\parallel C'B$, so $\angle XEY=\angle X'EY'=\angle DC'B$. Therefore, we have that $\angle XAY=\angle XEY=\angle BC'D=\pi-\angle XCD$, so we're done.

Fix $A$, $B$, and $D$. Then, we see that $C$ lies on a circle $\omega$ with circumcenter the point where $AE$ hits the circumcircle of $\triangle ABD$ (label this $O$). As a result, $C$ to $X$ and $Y$ is projective so we need $3$ points that work; note that $C$ being the point where $AB$ hits $\omega$ and the point where $AD$ hits $\omega$ results in two points that work because of right angles since for example if $A$, $B$, and $C$ are collinear, then $\angle ACD = 90^{\circ} - \angle CAO$ from the given condition and so $\angle OCA + \angle CAO = 90^{\circ}$ and thus $CO$ and $XY$ are parallel and hence we have a circle through the four points with curvature $0$, e.g. a line. Doing the same for $D$ and then letting $C=B$ and $C=D$ works so we are done.

I think my solution is slightly different from the ones above, so I'll sketch it out here.

Same as the other mmp but I wrote out the angle chases lol. Note that the incenter of $\triangle ABD$ must lie on $(CBD)$ by the angle condition. One may look at this the other way around, so let $C$ move on $(IBD).$ Then $X \mapsto{B} C \mapsto{D} Y,$ and as we want to prove $\measuredangle XAY=\measuredangle XCY,$ we show this map is just a rotation with angle $\measuredangle XCY$ about $A.$ It suffices to check this when $C=D,I,B.$ $C=I.$ Let $X'$ be the point where $ABEX'$ is concyclic. Note that $\angle X'BE=\angle EAX'=\angle AEX'=\angle X'EA,$ so $X'$ must lie on the perpendicular bisector of $AE.$ As there is only one point that is $\overline{BI} \cap \text{Perpendicular Bisector},$ this case is done. $C=B$ Let $\angle A=a,\angle B=b, \angle D=d.$ As $Y$ is on $\overline{BC},$ $YX$ is then the angle bisector of $\angle AYB,$ so $$\angle AYX=\angle XYE=90-\angle AEY=90-(\frac{a}{2}+b)=\frac{d}{2}-\frac{b}{2}$$and $$\angle ABX=\angle YBX-\angle ABY=\angle DGX-b=\frac{d}{2}+\frac{b}{2}-b=\frac{d}{2}-\frac{b}{2},$$so doing the same thing except $C=D,$ we're done.$\blacksquare$