Let $H$ be the orthocenter of triangle $ABC$. If $M$ is midpoint of $DY$, let $P$ be the reflection of $E$ with respect point $M$. Of course $BCPY$ is isosceles trapezoid. The bisector of $\angle BAC$ and perpendicular bisector of $BC$ intersect at $M$, so $M$ lies on circumcircle of triangle $ABC$. Thus we have $BD\cdot DC=AD \cdot DM$. Points $H$ and $O$ are isogonal cojugate wrt triangle, so easy angle chasing and we have that $\angle AXD= \angle DMP$. Thus quadrangle $AXMP$ is cyclic, so $DX\cdot DP=AD\cdot DM = BD\cdot DC$ - quadrangle $BSCP$ is cyclic, so points $B, X, C, P, T$ lie on the same circle.

Nice problem, nice idea for solution; however it could be made shorter: Clearly $\Delta ADX\sim\Delta AMO$, so $\Delta ADX$ is isosceles, and so is $\Delta DPM$, since $M$ is the center of the rectangle $DEYP$; as $\angle MPD=\angle MDP=\angle XDA=\angle XAD$ we infer $AXMP$ cyclic, and we are done. Best regards, sunken rock

Absolutely trivial for a G4. Here is the solution. Call Y' the reflection of Y through the perpendicular bisector of BC. Clearly if we show that BXCY' is cyclic, we're done. So if we show $DX(DY)=DB(DC)$, we're done. So if we show $DX(EY)=BD(DC)$, we're done. Assume $\angle B \textgreater \angle C$. Call $(A/2)+C=\delta$ Now, note that $\angle DAX = \angle DAC - \angle OAC = A/2 - 90 + B = \angle ADX = \delta$, so $DX = \frac{AD}{2sin \delta}$. Also, we can see that $EY=2DMtan\delta$ since $\angle EDY = \angle BDA = \delta$. So we wish to prove $DB(DC)=\frac{DM(DA)}{cos \delta}$. But call $Z$ the midpoint of arc $BC$ that does not contain $A$ in the circumcircle of $ABC$. We see that $DA(DZ)=DB(DC)$ and $DZ = \frac{DM}{cos \delta}$. So we're done. Horrible problem, nothing interesting about it. Very poor choice for G4, in my opinion.

I swear this has been posted a while ago, but I can't find it!!! my solution was the same as sunken rock's

You are right XmL here it is : http://www.artofproblemsolving.com/Forum/resources.php?c=114&cid=75&year=2013&sid=7c75a2addf93750e55d6db969f0c3964

lyukhson wrote: Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic. As shown in the figure below. $H$ is the intersection of $\circ O$ and $AY$. $F$ is the midpoint of $AD$. $DG//EY$ and $DG=EY$. It's obvious that $B,C,G,Y$ are concyclic. and $HE=HD$ so $H$ is the midpoint of $YD$. As $DF\times DY=\frac{AD}{2}\times DY=AD\times \frac{DY}{2}=AD\times DH=BD\times DC$ $B,F,C,Y$ are concyclic. Further, as $\triangle DXF \sim \triangle DYE$ we have $DF\times DY = DX\times EY = DX\times DG$ Thus $B,Y,G,C,F,X$ are concyclic.

Assume $AB < AC$ and let $L$ be the midpoint of arc $BC$. Denote by $Y'$ the reflection of $E$ over $L$; evidently $BY'YC$ is a isosceles (and hence cyclic) trapezoid. Now $\triangle DLY' \sim \triangle AXD$ because they are both isosceles (since $\triangle AXD \sim \triangle AOL$) and share an angle, so we see that $\angle XAL = \angle XAD = \angle DY'L = \angle XY'l$ and so $AXLY'$ is cyclic; hence $XD \cdot DY' = AD \cdot DL = BD \cdot DC$ so $XBY'C$ is cyclic as well, and we're done. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(13cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (0.64, 0.96); pair B = (0.4, -1.0); pair C = (3.0, -1.0); pair D = IntersectionPoint(Line(incenter(A,B,C),A,lisf),Line(B,C,lisf)); pair E = B+C-D; pair M = midpoint(B--C); pair O = circumcenter(A,B,C); pair X = IntersectionPoint(Line(D,D+O-M,lisf),Line(A,O,lisf)); pair Y = IntersectionPoint(Line((2)*(O)-X,E,lisf),Line(A,D,lisf)); pair L = midpoint(D--Y); pair Y_prime = (2)*(L)-E; /* Draw objects */ draw(A--B, rgb(0.2,0.6,0.0)); draw(B--C, rgb(0.2,0.6,0.0)); draw(C--A, rgb(0.2,0.6,0.0)); draw(circumcircle(A,B,C), rgb(0.0,0.4,0.0) + linewidth(1.0) + linetype("4 4")); draw(circumcircle(B,C,X), rgb(0.6,0.0,0.6)); draw(A--O, rgb(0.2,0.6,0.0)); draw(E--Y, rgb(0.6,0.0,0.6)); draw(Y_prime--L, rgb(0.6,0.0,0.6) + linetype("4 4")); draw(D--X, rgb(0.0,0.6,0.6)); draw(D--Y_prime, rgb(0.6,0.0,0.6) + linetype("4 4")); draw((abs(dot(unit(L-Y_prime),unit(D-Y_prime))) < 1/2011) ? rightanglemark(L,Y_prime,D) : anglemark(L,Y_prime,D), rgb(0.6,0.0,0.6)); draw((abs(dot(unit(D-A),unit(X-A))) < 1/2011) ? rightanglemark(D,A,X) : anglemark(D,A,X), rgb(0.6,0.0,0.6)); draw(A--D, rgb(0.2,0.6,0.0)); draw(D--Y, rgb(0.6,0.0,0.6)); draw(O--M, rgb(0.2,0.6,0.0)); draw(M--L, rgb(0.6,0.0,0.6)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(E); dot(M); dot(O); dot(X); dot(Y); dot(L); dot(Y_prime); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * dir(225)); label("$C$", C, lsf * dir(-45)); label("$D$", D, lsf * dir(45)); label("$E$", E, lsf * dir(45)); label("$M$", M, lsf * dir(45)); label("$O$", O, lsf * dir(45)); label("$X$", X, lsf * dir(60)); label("$Y$", Y, lsf * dir(-90)); label("$L$", L, lsf * dir(45)); label("$Y'$", Y_prime, lsf * dir(-90)); [/asy][/asy] In other news, 2048th post!

Let $F$ be a foot of perpendicular from $X$ to $AD$.Then it it easy to prove that triangles $DEY$ and $XFD$ are similar. So $EY/DY=EY/2DL=AD/2XD$ so $EY\cdot XD=DL\cdot AD=BD\cdot CD=CE\cdot CD$ ( we have obviously $BD=CE$) so triangles $CEY$ and $CDX$ are similar.Then $\angle XCY=90$, analogously $\angle XBY=90$ so done.

Sketch, will complete this soon: Let $M$ be the midpoint of $OM.$ Extend $OM$ to meet the circumcircle of $\triangle ABC$ at $J.$ It is easy to prove that $AJ$ bisects $\angle BAC,$ and as a consequence $A,D,J$ are collinear. Now, note that $\triangle ADX \sim \triangle DJM.$ By some easy ratio computations using PoP of $D$ wrt the circumcircle of $\triangle ABC,$ it follows that $\triangle BDX \sim \triangle BEY$ and similarly $\triangle CDX \sim \triangle CEY.$ The rest is simple angle chasing.

Let M be the midpoint of the minor arc BC and let Y' be the reflection of Y across the perpendicular bisector of BC.Now,it is easy to see that that BCYY' is an iscocelles trapezoid and X,D and Y' are collinear and M is the midpoint of DY(simple angle chasing).Now,by angle chasing we get <Y'MA=<AXY',so AXMY' is a cyclic.Also,BMCA is a cyclic so we have DB*DC=DM*DA=DX*DY',so we are finished.

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=530055 Sincerely Jean-Louis

Call the midpoint of $BC$ $M$ and $AD \cap \odot ABC = P$ Obviously $OMP$ is a straight line Consider reflection in line $OP$ as $R(OP)$. $R(OP) : D \to E, B \to C, O \to O, M \to M, P\to P, Y \to Y'$ $R(OP):DPY \to EPY'$ $\implies EPY'$ is a straight line. Also easily, $XDY'$ is a straight line. $\angle XAD = \angle \frac{A}{2}-90+ \angle B = 90-\angle C-\angle \frac{A}{2}$ $\angle PY'D = \angle EY'D = 90-\angle DEY' = 90-\angle EDP = 90 - \angle DBP - \angle DPB = 90-\angle C-\angle \frac{A}{2}$ $\implies \angle XAD=\angle PY'D$ $\implies XAY'P$ is cyclic $\implies DX \cdot DY'=DA \cdot DP$ $\implies DX \cdot DY'=DB \cdot DC$ $\implies BXCY'$ is cyclic. $\implies \angle BY'C = 180-\angle BXC$ But by $R(OP):\angle BY'C \to \angle CYB$ $\implies \angle CYB+\angle BXC=180$ $\implies BXCY$ is cyclic and thus we are done

Let $M$ be the midpoint of $\overline{BC}$ and let $T$ be the midpoint of arc $\widehat{BC}.$ Let $V$ be the projection of $X$ onto $AD$ and let $O_1$ be the midpoint of $\overline{XY}.$ Note that $(MT \parallel EY) \perp BC \implies \triangle DMT \sim \triangle DEY \implies \tfrac{DT}{DY} = \tfrac{DM}{DE} = \tfrac{1}{2}.$ Therefore, $T$ is the midpoint of $\overline{DX}.$ Meanwhile, $(XD \parallel OT) \perp BC \implies \triangle AXD \sim \triangle AOT \implies \triangle AXD$ is isoceles. Therefore, $V$ is the midpoint of $\overline{AD}.$ Then by Power of a Point, \[DV \cdot DY = DA \cdot DT = DB \cdot DC,\]implying that $B, C, V, Y$ are concyclic. Let $\omega \equiv \odot(BCVY).$ We claim that $O_1$ is the center of $\omega.$ Indeed as $O_1$ is the circumcenter of right triangle $\triangle XYV$, we have $O_1V = O_1Y.$ Meanwhile, as $MT$ is the $Y$-midline of $\triangle YXD$, it follows that $O_1 \in ML.$ Therefore, $O_1B = O_1C$ as well, so $O_1$ is the center of $\omega.$ Hence, $X \in \omega$, as desired.

Probably similar to other solutions...(I didn't read any of them so...) Assume WLOG, $AB<AC$. Let $V$ be the mid-point of $AD$,i.e., $V$ is the projection of $X$ onto $AD$.Let the tangent to the circumcircle of triangle $ABC$ at $A$ meet $BC$ at $T$. Note that $\angle TAD = \angle TDA = \frac{\angle A}{2}+\angle C$ and so $TA=TD$ and since $VX$ is the perpendicular bisector of $AD$, $T,V,X$ are col-linear. Now, note that by power of a point, $DB.DC = DA.DL = \frac{DA}{2}.(2DL) = DV.DY$ and therefore, $B,V,C,Y$ are concyclic. Now, $TD^2 = TA^2 = TB.TC = TV.TX$ and so points $B,V,X,C,Y,Y$ are con-cyclic.$\blacksquare$.

Let $AD$ hit the circumcircle of $\triangle ABC$ at $N$. Trivially, $N$ is the midpoint of $DY$. Set $Y'$ so that $EDY'Y$ is an rectangle. $N$ is the center of this rectangle. By symmetry, we have $\angle BYC = \angle BY'C$. It suffices to prove that $B,X,C,Y'$ are cyclic, or $BD \cdot DC = XD \cdot DY'$. Now we have $AD \cdot DN = BD \cdot DC$, so it suffices to prove $AD \cdot DN = XD \cdot DY'$, or $A, X, N, Y'$ cyclic. We angle chase $$\angle NY'X = \angle NY'D = \angle NDY' = \angle ADX = 90-\angle ADC = 90-(180-\angle DCA - \angle DAC) = 90-(180-\frac{1}{2} \angle A - \angle C)$$$$= \frac{1}{2} \angle A + \angle C - 90= 90 - \angle B - \frac{1}{2} \angle A= \angle OAC - \angle DAC= \angle XAC - \angle DAC = \angle XAD = \angle XAN$$so we have $A, X, N, Y'$ cyclic as desired. $\blacksquare$

A little simple for a G4, but a very neat problem nonetheless. Let the midpoint of $BC$ be $M$ and let the reflection of $Y$ wrt $OM$ be $Y'$. Then, simple angle chasing reveals that $\angle BYC = \angle BY'C$, so it is enough to show that $BXCY'$ is cyclic. Observe that $EY' \cap AY \equiv Z$ is the arc midpoint of $BC$, since it lies on $OM$ and $AD$. Therefore, $BD \cdot CD = AD \cdot ZD$. Furthermore, $XD \perp BC$, so $XD$ is parallel to $OZ$. Thus, $\angle EY'D = \angle EYD = \angle OZA$. Since $Z$ lies on the circumcircle of $OAZ$, $\angle OZA = \angle XDA = \angle XAD$. So $\angle EY'D = \angle XAD$, and $XAY'Z$ is cyclic. By PoP, we have $XD \cdot DY' = AD \cdot DZ = BD \cdot CD$, and $BXCY'$ is cyclic as desired.

Let $N,M,G$ be resp the midpoints of $AD,BC$ and arc $BC$ not containing $A$. we'll prove that both $BNCY;BNXC$ are cyclic. consider $\phi$ the composition inversion $\psi(A,\sqrt{AB.AC})$ and reflection around $AD$. clearly $\phi(N)$ is the reflection of $N$ wrt $G$,and $\phi(X)$ lays on the $A$-alttitude. also: $\angle ANX=\angle AX'N'=90\rightarrow GX'=GN'$ and $X'N'\parallel BC$. on the other hand $\angle X'GC=360-\angle AGX'-\angle AGC=360-2\angle ADB-\angle B=180-\angle C=BGN'$ i.e $\angle BGX'=N'GC$. therefore $\Delta BGX'\sim \Delta CGN'$, further more $BX'=N'C \rightarrow BCN'X'$ is an isosceles trapeziod, and thus cyclic. the latter implies that $BNXC$ is cyclic as well. now, $DY.DN=2DG.AD/2=DG.DA=DB.DC \rightarrow BNCY$ is cyclic. and the result follows.

D a m n, trig stays winning... Reduce the condition to $BD \times CD = XD \times YE$ Let $L$ be the arc midpoint, and $M$ the midpoint of BC then by homothety, $YE=2LM=a \tan(\frac{A}{2})$ Then find $XD$ in terms of $AD$ by sine rule in $\Delta AXD$ and do some simple trig and voila, we're done

Assume $AB<AC.$ Let $M$ be the midpoint of $DY.$ First, we have $XD \parallel EY,$ so $\angle ADX=\angle EYM.$ Next, since $OM \perp BC \perp XD$ we have $\angle AXD=\angle AOM=2 \angle ACM=2 (\angle ACB+\angle CAD) =2 \angle ADB=2 \angle EDM=\angle EMY,$ so $\triangle AXD \sim \triangle EMY$ and $\frac{XD}{AD}=\frac{MY}{EY}$ so $AD \cdot MY=XD \cdot EY.$ Since $ABMC$ is cyclic, we have $BD \cdot DC=AD \cdot DM=AD \cdot MY=XD \cdot EY.$ Now let $Y'$ be the reflection of $Y$ over the perpendicular bisector of $BC.$ We have $DY'=EY,$ so $BD \cdot DC=XD \cdot DY',$ and $BXCY'$ is cyclic. Clearly $BCYY'$ is cyclic as well, so $BXCY$ is cyclic.

Define the points as shown in the diagram, with Z the intersection of perp. through D to BC with (DEY); it's evident that the circumcenter of DEYZ is the midpoint of arc BC, and that BCYY' is cyclicislscelestrapezoid. The finish is as follows: $$ADX=FDZ,AO=OF\implies AX=XD,FD=FZ\implies AXD\sim DFZ\implies BD\cdot DC=AD\cdot DF=XD\cdot DZ\implies X\in(BCZ),$$and combining we have the desired. $\blacksquare$

Let the second intersection of $XD$ with $(BCY)$ be $T$. It is easy to see that $BTCY$ is an isosceles trapezoid, meaning it is cyclic. Let $N$ be the arc midpoint of $BC$. Notice that $DNT$ and $AXD$ are isosceles and have the same base angle, so they are similar. Then $\angle AXT = \angle AXD = \angle DNT = \angle ANT$, so $AXNT$ is cyclic. Note \[ XD \cdot TD = AD \cdot ND = BD \cdot CD, \]implying $BXCT$ is cyclic, which finishes.

Let $\overline{AD}$ intersect $(ABC)$ again at $S$, so $S$ is the midpoint of $\overline{DY}$. Let $N$ be the midpoint of $\overline{AD}$. By power of a point, $BCYN$ is cyclic. Let $Y'$ be the reflection of $Y$ over the perpendicular bisector of $\overline{BC}$, which also lies on $BCYN$. Clearly $X,D,Y'$ are collinear, hence $\angle XY'Y=90^\circ$. On the other hand, since $OA=OS$, by a homothety at $A$ we also have $XA=XD$, hence $\angle XND=\angle XNY=90^\circ$, so $BCXYNY'$ is cyclic. $\blacksquare$

Let $P$ be the reflection of $Y$ across $M$ and let $F$ denote the midpoint of arc $BC$. Also let $M$ be the midpoint of $BC$. Noting that $\angle EYD = \angle 90 - \angle YDE = 90 - \angle EFD = \angle FEY$ we have that $DEPY$ is a rectangle with $F$ being the midpoint of $EP$ and $DY$. It is also clear that $P \in (BYC)$. Claim: $\triangle AXD \sim \triangle EFY$ Proof: This is just angle chasing. Note that $\angle OAF = \angle OFD = \angle OFE = 90 - \angle DEF = \angle FEY$. Also $\angle XDA = \angle YDP = \angle DYE$. Thus proved. Then we clearly have $XD \cdot DP = AD \cdot DF = BD \cdot BC$ so we're done.

Solved with emotional support from Megarnie, Tienxion, Redfiretruck, and OronSH Define $X'$ and $Y'$ as the reflections of $X$ and $Y$ across the perpendicular bisector of $BC$. Note that $XX'BC$ is a Cyclic Isosceles Trapezoid $YY'BC$ is a Cyclic Isosceles Trapezoid $XX'YY'$ is a Cyclic Rectangle Now note that $XX' \parallel YY' \parallel BC$ hence the pairwise radical axes of $(XX'BC)$, $(YY'BC)$ and $(XX'YY')$ don't concur. Which means $(XX'BYY'C)$ cyclic which means $(BXCY)$ is cyclic $\blacksquare$

Quite boring stuff. Let $M'$ be the arc midpoint of minor arc $BC$. Let $H$ be the foot of the altitude from $A$. Now, note that $$AH \parallel XD \parallel MM' \parallel EY$$Notice that $$\alpha = \angle DAX = \angle DAC - \angle OAC = \angle DAB - \angle BAH=\angle HAD$$($\angle BAH = \angle CAO = 90 - \angle B$). But, then via the parallel lines $$\alpha = \angle HAD = \angle XDA = \angle MM'D = \angle EYD$$Now, notice that $$\angle DM'M=\angle EM'M $$via reflection and in turn this means $$\angle AXD = 180 - 2\alpha = 180 - 2\angle DM'M = 180 - \angle DM'M - \angle MM'E = \angle EM'Y$$Thus, we must have that $\triangle AXD \sim \triangle EM'Y$. Now, notice the following. Since $M$ is the midpoint of $DE$ and $MM'$ is parallel to $EY$ , we have that $M'$ is the midpoint of $DY$. By the previous similarity, we have that $$\frac{AD}{EY}=\frac{XD}{M'Y}\implies AD \cdot M'Y = XD \cdot EY \implies AD \cdot DM' = XD \cdot EY$$So, \begin{align*} BE \cdot BD &= BD (BM + EM)\\ &= BD (CM + DM)\\ &= BD \cdot CD \\ &= AD \cdot DM'\\ &= XD \cdot EY \text{ (via the previous similarity)} \end{align*}But then, this means that $\triangle XBD \sim \triangle EBY$ which in turn implies that we must have $\angle XBY = 90^\circ$. Now, let $K$ be the intersection of $MM'$ and $XY$. Since $M'$ is the midpoint of $DY$ and $XD \parallel MK$, $K$ must also be the midpoint of $XY$. Thus, $K$ is the center of $\triangle BXY$ ($\angle XBY = 90^\circ$) and $BK = XK = KY$. Further, $K$ lies on the perpendicular bisector of $BC$ ($MM'$) and so it is equidistant to $B$ and $C$. Thus, $BK=CK$ which gives us that $$XK = KY = BK = CK$$and thus $K$ must be the center of $(XBYC)$ and thus $XBYC$ is indeed cyclic.

Let $M = AD \cap (ABC)$ and let $Y'$ be the reflection of $Y$ wrt $OM$. Then $BCYY'$ is isosceles trapezoid, so it suffices to prove that $BXCY'$ is cyclic. Let $A' = Y'M \cap (ABC)$ and let $N'$ be the midpoint of $AE$. It's not hard to see that $MD = ME = MY'$, so we have $Y'E \cdot EN' = ME \cdot A'E = AD \cdot DM = BD \cdot DC = BE \cdot EC$, therefore $BY'CN'$ is cyclic. Thus it's enough to prove that $BCN'X$ is cyclic. Let $N$ be the midpoint of $AD$. Then $BCNN'$ is isosceles trapezoid, so it suffices to prove that $BCXN$ is cyclic. Let $T = NX \cap BC$. Simple angle chasing gives $AX = XD$, thus $XN$ is the perpendicular bisector of $AD$, hence $TA = TD$. Therefore $AT$ is tangent of $(ABC)$, so $TB \cdot TC = TA^2 = TN \cdot TX$, so we're done. $\blacksquare$

Easy for a G4(Still spent an hour on it) Basically we have to prove XDB is similar to BEY. This can be done using trig observations

We will show that $\angle XBY=90$, which is of course sufficent by symmetry. Let $F$ be the foot from $A$ to $BC$, and let $L$ be the midpoint of arc $BC$. Note that since $BD\perp DX$ and $BE\perp EY$, the perpendicularity we wish to show is equivalent to $$BD\cdot BE=XD\cdot YE.$$ We can compute three of the lengths relatively easily. By angle bisector theorem, $$BD=\frac{ac}{b+c},BE=\frac{ab}{b+c},$$and $$YE=2ML=a\tan\alpha/2.$$Thus, the desired conclusion is equivalent to $$XD=\frac{abc}{(b+c)^2\tan\alpha/2}.$$ The idea is to use similar triangles $\triangle ADX$ and $\triangle ALO$. Due to this, $$XD=R\cdot\frac{AD}{AL}=R\cdot\frac{AF}{AF+ML}=R\cdot \frac{\frac{bc}{2R}}{\frac{bc}{2R}+\frac{a}{2}\tan\alpha/2}$$$$=\frac{bcR}{bc+aR\tan\alpha/2}.$$ Thus, it suffices to show that $$\frac{abc}{(b+c)^2\tan\alpha/2}=\frac{bcR}{bc+aR\tan\alpha/2}$$$$abc+a^2R\tan\alpha/2=R(b+c)^2\tan\alpha/2$$$$abc=R\tan\alpha/2((b+c)^2-a^2).$$Let $A$ be the area of the triangle. Then, $abc=4RA$, so this becomes $$4A=\tan\alpha/2(b+c-a)(b+c+a).$$Squaring and using Heron's formula, this is equivalent to $$\tan^2\alpha/2=\frac{(a-b+c)(a+b-c)}{(a+b+c)(-a+b+c)}.$$However, we have $$\tan^2\alpha/2=\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}=\frac{a^2-b^2-c^2+2bc}{2bc+b^2+c^2-a^2}=\frac{(a-b+c)(a+b-c)}{(a+b+c)(-a+b+c)},$$as desired.

Let $N$ denote the midpoint of minor arc $\hat{BC}$, and let $Z$ denote the reflection of $Y$ over $MN$. Observe that $AXNZ$ is cyclic. Indeed, \[\measuredangle XAD = \measuredangle OAN = \measuredangle ANO = \measuredangle DYE = \measuredangle DZE.\]Also similarly observe that $CYZB$ is trivially cyclic. Hence it now suffices to show $BZCX$ is cylic, which is trivial as \[DX \cdot DZ = DA \cdot DN = DB \cdot DC, \]as desired. $\blacksquare$

Letting $M$ be the midpoint of arc $BC$, we have $XD \parallel OM$, so $\triangle AXD \sim \triangle AOM$. Therefore, $\triangle AXD$ is isosceles. Let $F$ be the midpoint of $\overline{AD}$. By power of a point, $FBYC$ is cyclic. Because $X$ and $Y$ are equidistant from line $OM$, it follows that the circumcenter of right triangle $XFY$ lies on $OM$. So, $(XFY)$ must intersect line $BC$ at two points equidistant from $O$. For power of a point reasons, the only circle passing through $F$ and $Y$ that satisfies the aforementioned property is $(BFCY)$; therefore, $(BFXCY)$ is cyclic.

Easy for a G4 [asy][asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; real xmin = -7, xmax = 8, ymin = -5, ymax = 9; /* image dimensions */ draw((0.9282707692837183,5.124674700939942)--(-0.2,0)--(7,0)--cycle, linewidth(0.7)); /* draw figures */ draw(circle((3.4,1.893948134588014), 4.067805248104857), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(-0.2,0), linewidth(0.7)); draw((-0.2,0)--(7,0), linewidth(0.7)); draw((7,0)--(0.9282707692837183,5.124674700939942), linewidth(0.7)); draw(circle((3.4,-0.7457494356122296), 3.676430635917937), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(4.136200965821236,0), linewidth(0.7)); draw((-5.770044925854284,0)--(0.9282707692837183,5.124674700939942), linewidth(0.7)); draw((-5.770044925854284,0)--(2.663799034178764,2.856215355809231), linewidth(0.7)); draw((-5.770044925854284,0)--(-0.2,0), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(4.136200965821236,-4.34771422703369), linewidth(0.7)); draw((2.663799034178764,2.856215355809231)--(2.663799034178764,-4.3477142270336895), linewidth(0.7)); draw((4.136200965821236,0)--(4.136200965821236,-4.34771422703369), linewidth(0.7)); draw((3.4,1.8939481345880143)--(3.4,-2.1738571135168425), linewidth(0.7)); dot((0.9282707692837183,5.124674700939942),linewidth(3pt) + dotstyle); label("$A$", (0.3859373501577617,5.175259523212921), NE * labelscalefactor); dot((-0.2,0),linewidth(3pt) + dotstyle); label("$B$", (-0.759590777205151,-0.68818809213788084), NE * labelscalefactor); dot((7,0),linewidth(3pt) + dotstyle); label("$C$", (7.2362103849019395,-0.30079460486524406), NE * labelscalefactor); dot((3.4,1.8939481345880143),linewidth(3pt) + dotstyle); label("$O$", (3.488340639449198,2.010391738163943), NE * labelscalefactor); dot((2.6637990341787634,0),linewidth(3pt) + dotstyle); label("$D$", (2.1765862285407382,-0.7090095907237294), NE * labelscalefactor); dot((3.4,0),linewidth(3pt) + dotstyle); label("$M$", (3.238482656419015,-0.6298310893095779), NE * labelscalefactor); dot((4.136200965821236,0),linewidth(3pt) + dotstyle); label("$E$", (4.362843580054838,-0.5298310893095779), NE * labelscalefactor); dot((2.663799034178764,2.856215355809231),linewidth(3pt) + dotstyle); label("$X$", (2.6138376988435583,3.0306451688705214), NE * labelscalefactor); dot((4.136200965821236,-4.34771422703369),linewidth(3pt) + dotstyle); label("$Y$", (4.446129574398232,-4.735773803650981), NE * labelscalefactor); dot((2.663799034178764,-4.3477142270336895),linewidth(3pt) + dotstyle); label("$Y'$", (1.9,-4.87), NE * labelscalefactor); dot((3.4,-2.1738571135168425),linewidth(3pt) + dotstyle); label("$M_{A}$", (2.678031748029836,-2.86239450661285), NE * labelscalefactor); dot((1.7960349017312414,2.56233735046997),linewidth(3pt) + dotstyle); label("$S$", (1.28126178934925,2.6350366957393994), NE * labelscalefactor); dot((-5.770044925854284,0),linewidth(3pt) + dotstyle); label("$R$", (-6.235299200142082,-0.6339385822388206), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $M_A$ be the arc-midpoint of minor-arc $\widehat{BC}$ and since $\overline{O-M-M_A}\parallel EY$, due to MPT it is also the midpoint of $DY$. Let the $AA\cap BC=R$ so it is well-known that $TX$ is the perpendicular bisector of $AD$, suppose that point is $S$, by POP we have \begin{align*} RS\cdot RX=RA^2=RB\cdot RC\implies \odot(BSXC)\\ BD\cdot CD=AD\cdot M_AD=SD\cdot YD\implies \odot(BSCY) \end{align*}Thus, $\odot(BXCY)$ is cyclic.

Let $M$ be the midpoint of $BC$, $M_a$ be the midpoint of arc $BC$ not including $A$ on $(ABC)$. Since $M_a$ is the midpoint of $DE$, we have that $DM=MY$. Let $Y'$ be the reflection of $Y$ across the perpendicular bisector of $BC$. Note that $(BCYY')$ are concyclic, $DY' \perp BC$, and $DY' = 2MM_a$. Note that the exists a circle centred at $X$, passing through $A,D$, and tangent to $(ABC)$ and $BC$. Call this circle $\omega$. Let $D'$ be the antipode of $D$ in $\omega$. Now, $\angle DAD' = 90^\circ = \angle MM_aD$ and $\angle ADD' = \angle M_aMD$ since $DD'$ and $DM$ are both perpendicular to $BC$. So we have $\Delta ADD' \sim M_aMD$, so $\frac{DD'}{MD} = \frac{AD}{MM_a} \implies DD' \cdot MM_a = DA \cdot DM$. By power of point, $DY' \cdot DX = 2MM_a \cdot \frac{DD'}{2} = DD' \cdot MM_a = DA \cdot DM = DB \cdot DC$, so $(BXCY')$ concyclic. It follows that $(BXCY)$ concyclic, so we are done.