It's still nice to have a solution posted . Let $F$ be the point on the circle $(ABC)$ s.t. $BF\|AY$. We see that the four lines $BF,BZ,BA,BE$ for a harmonic quadruple, so their intersections with the circle (other than $B$) for a harmonic quadrilateral, i.e. the quadrilateral $AFEZ$ is harmonic. This means that the tangents from $A,E$ to the circle meet on the diagonal $ZF$, so $D,Z,F$ are collinear. We will keep this in mind in what follows. Since $\angle ZAB=\angle BFD$ and $\angle AZB=\angle ACB=\angle FBD$, we find the triangles $ZAB,BFD$ to be similar. At the same time, the triangles $ZCD,BFD$ are similar, so $ZAB,ZCD$ are similar. What we have to prove is now reduced to a simple angle chase: $\angle CDZ=\angle ZBA=\angle ZAD$, and that's it.

Cool, thanks. Btw, where can I find more information on harmonic quadrilaterals? I tried searching on google, but that didn't bring up much.

Sorry.. I could find next to nothing. I remember seeing long thread on Math Forum once, but couldn't find it. I'll keep looking for info. [Edit: try looking for stuff in this thread. It's not about harminic quadrilaterals per se, but I think there's quite a bit of useful information on the subject in those messages]

It's somewhere in the 2004 or 2005 issues of http://www.cms.math.ca/Competitions/MOCP/ [Moderator edit: Indeed, it's Olymon 2004 problem #300, solved at http://www.cms.math.ca/Competitions/MOCP/2004/sol_mar.html and at http://www.kalva.demon.co.uk/short/soln/sh98g8.html .] I remember I had an algebraic solution---coordinates! lol

not important which one of B and C are bigger in my solution B is bigger one. let $ AE\cap BC\equiv H\ ,AZ\cap BC\equiv M$ we have $ \triangle ABX\sim\triangle ADH$ and also $ \angle XBY=\angle ZAH$ hence AZ bisect DH and cause $ (DH,BC)=-1$ hence $ MD^{2}=MB.MC=MZ.MA$

Let $ O$ be the midpoint of $ [BC]$ and let $ F$ be the orthogonal projection of $ A$ in $ BD$. So we have $ O\hat{F}A = O\hat{A}D = 90^{\circ}$, which means $ D$ and $ F$ are inverses with respect to the circumcircle of $ [ABC]$. So invert through the circumcircle of $ [ABC]$. Line $ BD$ is sent to itself and the circumcircle of $ [ADZ]$ is sent to the circumcircle of $ [AFZ]$. Therefore it suffices to prove line $ BD$ is tangent to the circumcircle of $ [AFZ]$. So we just have to prove $ AZ\perp FZ$. Now use complex numbers. Let $ a,b,c,e,f,x,y,z$ be the complex numbers corresponding to $ A,B,C,E,F,X,Y,Z$, respectively. Suppose $ b = 1$ and $ c = - 1$. So $ a,e,z$ are on the unit circle and $ f$ is on the real axis. Therefore we have: $ e = \bar{a} = \frac {1}{a}$ $ f = \frac {1}{2}(b + c + a - bc\bar{a}) = \frac {a^2 + 1}{2}$ $ \bar{f} = f = \frac {a^2 + 1}{2}$ $ x = \frac {1}{2}(b + e + a - be\bar{a}) = \frac {a^3 + a^2 + a - 1}{2a^2}$ $ y = \frac {a + x}{2} = \frac {3a^3 + a^2 + a - 1}{4a^2}$ $ \bar{y} = \frac {3 + a + a^2 - a^3}{4a}$ Since $ z$ belongs to line $ by$: $ \frac {y - b}{\bar{y} - \bar{b}} = \frac {z - b}{\bar{z} - \bar{b}}\Leftrightarrow\frac {y - b}{\bar{y} - \bar{b}} = - bz\Leftrightarrow z = - \frac {y - 1}{\bar{y} - 1}$, which means $ z = \frac {3a^2 + 1}{a(a^2 + 3)}$ and $ \bar{z} = \frac {a(a^2 + 3)}{3a^2 + 1}$. Now we want to prove $ az\perp fz$: $ \frac {z - f}{\bar{z} - \bar{f}} = - \frac {z - a}{\bar{z} - \bar{a}}\Leftrightarrow\frac {z - f}{\bar{z} - \bar{f}} = - ( - az)\Leftrightarrow f - z = a(fz - 1)$, and it's trivial to check the last equality.

Dear Mathlinkers, for the beauty of this nice result, can we imagine a proof without calculation ? I think so. Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, for the beauty of this nice result, can we imagine a proof without calculation ? I think so. Sincerely Jean-Louis I think that bouth solutions are interesting, in fact we must lear to solve problems by bouth ways

Dear Raül and Mathlinkers, you are right... all solutions of a problem is interesting for all members of our community... after each one of us has his proper sensibility. Do you have find yours? Sincerely Jean-Louis

I found one mine yes this problem was used for one friendly competition of geometry (in witch each participant propose 2 problems) and when I find out that it is a G8... xD My solution is the same as one presented here, is like that: We create the following points: $Q=AE\cap BC$, $P=A\cap BC$, $E'=DZ\cap C(ABC)/\{Z\}$ and $U=AE'\cap BC$ (we will find out that this point doesn't exist, by now assume that it exists and we goal one contradiction) where $C(ABC)$ is the circuncircle of $ABC$ We have the similarity of the triangles $[AXB]\sim [DQA]$ and $[AYB]\sim [DZA]$ so, since $\overline{AX}=2\overline{AY}$ we can conclude $\overline{DQ}=2\overline{DP}$ and so $P$ is the midpoint of $[DQ]$. By one well known fact (we know that $(A, E, Z, E')$ is one harmonic conjugate, just take the pencil from $A$ and intersect with $BC$) we get the points $(D, Q, P, U)$ are in harmonic division, but since $\overline{DP}=\overline{PQ}$ we find that the point $U$ must be in the infinity, and $AE'\parallel BC$, and now just angle chase with inscribed arcs: $\angle DAZ+\angle ZAC=\angle DBC=\angle CBA=\angle E'CB=\angle E'DB+\angle ZE'C=\angle ZDC+\angle ZAC\Leftrightarrow \angle DAZ=\angle ZDC$ witch is our goal

let DB intersect AE at point H and let M is midpoint of DH , let MA intersect BY at point Z' easy to see that trangle ABX ~ DAH and AM and BY are their medians so angle AZ'B = 180 - BAD so Z' is on circle arround ABC so Z' =Z and M is on radical line of point D and circle arround ABC so MZ*MA = MD*MD so done

Dear Mathlinkers, a some up of my synthetic proof. 1. T is the second point of intersection of DZ with the circumcircle (omega). 2. If we prove that AT // BC we are done. Who want to continue? Sincerely Jean-Louis

Here's another inversive proof Invert with center $D$ and radius $DA$.This takes $C$ to $B$,$B$ to $C$ and leaves $A,E$ Invariant.Let $Z'$ be the intersection of $\omega$ with $DZ$.This $Z'$ is the inverse of $Z$,So the line $AZ'$ is the inverse of the circumcircle of $\bigtriangleup ADZ$.So it suffices to prove that $AZ'||CB$. Taking $\omega$ as the unit circle,This is same as $z'=-\frac{b^2}{a}$. Some calculations(not too long...finding out the value of $x=\frac{a^2b+ab^2+a^3-b^3}{2a^2}$,then of $y=\frac{a^2b+ab^2+3a^3-b^3}{4a^2}$,then of $z=\frac{b^2(3a^2+b^2)}{a(3b^2+a^2)}$ and finally of $z'$) prove this result

The notations are all as in the picture. Lemma:- Very easy to see that the circle centered at E(the midpoint of BD) passing through D contains the point I An inversion with respect to G(the circumcenter of $ \Delta ABC $) with power $ R^2 $ where $ R $ is the circumradius of $ \Delta ABC $ will take $ \odot BDI $ to a circle passing through B and I and tangent to AL. BC is taken to $ \odot GBC $ by this inversion. So the tangency point is the diametrically opposite point of G in this circle which is G. So $ \odot BIL $ is tangent to $ AL $.

Well, this one is simple from angle chasing, no need to overkill such a nice one! Denote $F=AE\cap BC,$ so that $F$ is the midpoint of $AE,$ and therefore from the midpoint theorem we get $FY\parallel EX.$ So we obtain that $\angle YFA=\angle BEA=\angle YZA,$ ie $YFZA$ is a cyclic quadrilateral. Since $\angle AYF=\frac{\pi}{2},$ therefore $AF$ is the diameter of $\odot(YFZA).$ So we see that $AF\perp BD\implies FD$ is tangent to $\odot(YFZA).$ Now, note that $DA=DE$ so obviously $DE$ is tangent to $\odot(ABC),$ leading to $\angle ZBE=\angle ZAE;$ and since $DF$ is tangent to $\odot(YFZA),$ we also have $\angle ZAF=\angle ZFD.$ Considering these, we see that $\angle ZFD=\angle ZBD\implies ZDEF$ is cyclic. This time, we get that $\angle ZDF=\angle ZEF=\angle ZAD$ (since $DA$ is tangent to $\odot(ABC).$) And, we finish our proof here. $\Box$

Let $AE\cap BC = H$. $HY$ is midline of $\triangle EAX$. Let $\angle ABC = \alpha$. Then $\angle BAX = \pi /2 - \alpha$, $\angle HAY = \alpha$, $\angle AHY = \pi /2 - \alpha$, $\angle AYH = \pi/2$. Let $\angle CBZ = \beta$. Then $\angle ABZ = \alpha - \beta = \angle DAZ$, $\angle ZAC = \beta$, $\angle ADC = \pi /2 - 2 \alpha$. $\angle AHY = \angle ACB = \angle AZB$, so $AYHZ$ is cyclic. $\angle HAY = \angle BHY$, so $BH$ is tangent to the circumcircle of $AYHZ$. $AZ$ is radical axis of $(AZD)$ and $(AYHZ)$, so it is enough to prove that $AZ$ divides $[DH]$ in its middle, which is equivalent to $\frac{sin \angle DAC}{sin \angle ADC} = sin \angle ZAH = \frac{sin(\alpha-\beta)}{cos2\alpha}=sin(\alpha + \beta)$. $(*)$ Applying Ceva's theorem to $\triangle AHB$ and point $Y$ we get $\frac{sin(\alpha - \beta)}{cos2\alpha}*\frac{sin\alpha}{cos\alpha}*\frac{sin\alpha}{sin\beta} = 1$. $(**)$. Both $(*)$ and $(**)$ are equivalent to $(sin\alpha)^3 sin\beta = (sin\beta)^3 sin\alpha$. So $(*)$ follows from $(**)$. $QED$.

Let $R$ be the foot from $A$ to $BC$. Let $O$ be the midpoint of $BC$. As $R$ is the midpoint of $AE$, thus $YR\parallel XE$. Note that \begin{align*} \measuredangle AZY=\measuredangle AZB=\measuredangle AEX=\measuredangle ARY \end{align*}and because $\measuredangle RYA=\measuredangle EXA=90^\circ$ and $\measuredangle CRA=90^\circ$, we get that $(ARYZ)$ is tangent to $BC$. Let $S=DZ\cap (ABC)$. Invert at $D$ with radius $DA$, note that $(ARZ)$ maps to $(AOS)$, hence $BC$ is tangent to $(AOS)$. However because $\triangle AOS$ is isosceles, we get that $AS\parallel BC$, inverting back we get that $(AZD)$ is tangent to $BD$. We are done. [asy][asy]import geometry; size(10cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,E,X,Y,D,R,Z,S; O=(0,0);C=dir(0);B=dir(180);A=dir(55);path w=circumcircle(A,B,C);R=foot(A,B,C);E=2R-A;X=foot(A,B,E);Y=midpoint(A--X); Z=intersectionpoints(w,B--100Y-99B)[0];D=intersectionpoint(line(B,C),perpendicular(A,line(A,O))); S=intersectionpoints(w,D--100Z-99D)[1]; draw(A--B--C--cycle,deep);draw(w,heavyblue);draw(circumcircle(A,R,Z),heavyblue);draw(A--E,deep);draw(C--E--B,deep); draw(A--X,deep);draw(C--D,deep);draw(S--D,deep);draw(D--A,deep);draw(B--S--A,deep);draw(B--Z,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(270)); dot("$R$",R,dir(R)); dot("$E$",E,dir(E)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$D$",D,dir(D)); dot("$S$",S,dir(S)); [/asy][/asy]

Let $BC \cap AE=F$ It's trivial to see that $E \in \odot(ABC)$ as $A$ is the orthocentre. From angle chase we get,$\measuredangle ABF=\measuredangle EBF=\measuredangle XAF=\measuredangle EAC=\measuredangle CEA \Rightarrow EC||AX.$ Notice that $DE$ is tangent to $\odot(ABC)$ Now draw a pont $K$ such that $AX||BK||EC$ Let $AX \cap BK=P_{\infty}$ $$-1=(P_{\infty},Y;A,X)\stackrel{B}{=}(K,Z;A,E) \Rightarrow AKEZ \text {is a Harmonic quadrilateral}$$It suffices to show that $DKFE$ is cyclic. This is just simple angle chasing, $\measuredangle KBE=\measuredangle KZE=\measuredangle DZE=\measuredangle CFE=90^\circ \Rightarrow$ $DKFE$ is cyclic As $AD$ is tangent to $\odot(ABC)$, $\measuredangle DAZ= \measuredangle ZEC= \measuredangle EDZ $ which implies $BD$ is tangent to $\odot(DAZ)$ where the last equality came from cyclic quadrilateral.

Let $\overline{BC}$ and $\overline{AE}$ intersect at $M$. Since $\overline{BE} \parallel \overline{MY}$, $AYMZ$ is cyclic by Reim's. Notice that $\angle MAC=\angle AEC=\angle CAD$, so $C$ lies on the angle bisector of $\angle MAD$. Since $\angle BAC=90^\circ$, $B$ lies on the external angle bisector of $\angle MAD$. Since $Z$ lies on the circumcircle of $\triangle AMY$ and on the Apollonian circle of $\overline{BC}$ passing through $A$, $Z$ is the $A$-humpty point of $\triangle AMD$. Thus, the circumcircle of $\triangle ADZ$ is tangent to $\overline{BD}$. \[\mathcal{YUH}\]

Let the line $AE$ intersect $BC$ at $J$. Note that $Y$ and $J$ are the midpoints of $AX$ and $AE$ respectively, hence $YJ//XE$ This implies that $\angle AYJ= 90$ Also, since $\angle BJA= \angle AYJ= 90$ it means that $ \angle YJB= \angle YAJ$ Hence, $BC$ tangent to $AYJ$ $\angle AZB= \angle AEX= \angle AJY$ , so $A,Y,J,Z$ concyclic. Also, note $\angle ZED= \angle EAZ= \angle ZJD$ which implies that $J,Z,D,E$ concyclic Therefore, $\angle ZDC=\angle ZEA=\angle ZBA= \angle ZAD$ Hence, $BC$ is tangent to $ZAD$

Note that $\angle BAX=90^\circ-\angle ABX=90^\circ-2\angle ABC$ and that \[\angle ADC=\angle ACB-\angle CAD=\angle ACB-\angle BAC=90^\circ-2\angle ABC\]so $\angle BAX=\angle ADC.$ $~$ Let $F,G$ be intersections of $BC$ with $AE$ and $AZ$ respectively. $AF\perp BD$ so $\triangle BAX\sim \triangle ADF.$ Since $\angle FAG=\angle EAZ=\angle EBZ=\angle XBY$ so $G$ is midpoint of $FD.$ Since $GC\cdot GB=GZ\cdot GA$ It remains to show that $GC\cdot GB=GF^2.$ $~$ Note that \begin{align*} &= FB\cdot FC \\ &= FA^2 \\ &= AD^2 - FD^2 \\ &= DC\cdot DB - FD^2 \\ &= (DF+FB)(DF-FC) - FD^2 \\ &= DF(FB-FC) - FB\cdot FC \\ \end{align*}so $FB\cdot FC = \tfrac12 DF(FB-FC)=GF(FB-FC)$ and $GC\cdot GB = (GF+FB)(GF-FC)=GF^2+GF(FB-FC)-FB\cdot FC=GF^2$ as desired.

By symmetry $DE$ is tangent to $\omega$ at $E.$ Constructing $E-\text{antipode}$ $F$ we get $AF\parallel BC$ and $$BF\parallel AX\implies (AEZF)_{\omega}\stackrel{B}{=} (AXY\infty_{AX})=-1\implies D\in FZ.$$Finally $\measuredangle DAZ$ $=$ $\measuredangle AFD$ $=$ $\measuredangle BDZ$ yields the conclusion.

Let $A_1$ lie on $(ABC)$ such that $AA_1 \parallel BC$, $M = AE \cap BC$, and $T = AZ \cap BD$. Because $$AA_1 \parallel BC \perp AE$$we know $A_1E$ is a diameter of $(ABC)$. Thus, $$BA_1 \perp BE \perp AX$$implies $BA_1 \parallel AX$, which means $$-1 = (A, X; Y, \infty_{AX}) \overset{B}{=} (A, E; Z, A_1) \overset{A}{=} (D, M; T, \infty_{BC})$$so $T$ is the midpoint of $DM$. It's clear that $ABEC$ is a cyclic kite, so $$-1 = (A, E; B, C) \overset{A}{=} (D, M; B, C).$$It follows that $B$ and $C$ are inverses wrt $(DM)$, yielding $$TD^2 = TB \cdot TC = | Pow_{(ABC)}(T) | = TA \cdot TZ$$which finishes. $\blacksquare$ Remarks: It's very easy to reverse engineer the fact that $T$ is the midpoint of $DM$, as $-1 = (D, M; B, C)$ is well-known and commonly seen. At this point, the definition of $D$ motivates us to use projective methods to show the aforementioned result, and the rest follows naturally.

Let W be the antipode of E, $V=AE\cap BC$, and $M=AZ\cap BC$. Then \[WBE=90=AXE\Rightarrow AX\parallel BW\implies(DMVP_{\infty})\stackrel{A}{=}(AZEW)\stackrel{B}{=}(AYXP_{\infty})=-1,\]so M is the midpoint of DV. It follows that $M(DBVC)=-1\implies MD^2=MD\cdot MV=MB\cdot MC=MZ\cdot MA$, so $MD\equiv BC$ is tangent to (ADZ), as needed.

How is this G8? [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 = -15.98006838884981, xmax = 9.683036625251171, ymin = -10.19069829689637, ymax = 5.434389088786898; /* image dimensions */ /* draw figures */ draw((-1.0388665184850865,4.327988392401811)--(-5.374150879830733,-2.3104157859087104), linewidth(0.4)); draw((xmin, -0.6530612244897959*xmin + 3.649544951758489)--(xmax, -0.6530612244897959*xmax + 3.649544951758489), linewidth(0.4)); /* line */ draw((-5.374150879830733,-2.3104157859087104)--(9.338330097801927,-2.4489563366019524), linewidth(0.4)); draw(circle((1.982089608985597,-2.379686061255332), 7.356566624475532), linewidth(0.4)); draw((xmin, 0.4503730985071294*xmin + 4.795865925267252)--(xmax, 0.4503730985071294*xmax + 4.795865925267252), linewidth(0.4)); /* line */ draw((xmin, -0.00941653218813773*xmin-2.3610216506525457)--(xmax, -0.00941653218813773*xmax-2.3610216506525457), linewidth(0.4)); /* line */ draw((-15.56556976958695,-2.2144479618905266)--(-1.1646456874358049,-9.029282250092894), linewidth(0.4)); draw((xmin, -1.59611787065206*xmin-10.888194044787035)--(xmax, -1.59611787065206*xmax-10.888194044787035), linewidth(0.4)); /* line */ draw((xmin, 0.6265201451516054*xmin + 4.97885919435623)--(xmax, 0.6265201451516054*xmax + 4.97885919435623), linewidth(0.4)); /* line */ draw((-4.088852374348884,2.417110811275681)--(-5.374150879830733,-2.3104157859087104), linewidth(0.4)); draw(circle((-15.386643015476231,16.786894627489083), 19.00218500547524), linewidth(0.4)); draw((xmin, 0.6265201451516054*xmin + 1.0565980033894682)--(xmax, 0.6265201451516054*xmax + 1.0565980033894682), linewidth(0.4)); /* line */ draw((-4.396583103819279,1.285229616255652)--(-1.1646456874358049,-9.029282250092894), linewidth(0.4)); draw((-1.1646456874358049,-9.029282250092894)--(5.128824905406998,4.2699101275822295), linewidth(0.4)); draw((5.128824905406998,4.2699101275822295)--(-1.0388665184850865,4.327988392401811), linewidth(0.4)); draw((-1.0388665184850865,4.327988392401811)--(-4.396583103819279,1.285229616255652), linewidth(0.4)); draw((-15.56556976958695,-2.2144479618905266)--(5.128824905406998,4.2699101275822295), linewidth(0.4)); /* dots and labels */ dot((-1.0388665184850865,4.327988392401811),dotstyle); label("$A$", (-0.9485480942903854,4.553784452888563), NE * labelscalefactor); dot((-5.374150879830733,-2.3104157859087104),dotstyle); label("$B$", (-5.283832455636032,-2.084619725421958), NE * labelscalefactor); dot((9.338330097801927,-2.4489563366019524),dotstyle); label("$C$", (9.438070688100227,-2.2200973617140094), NE * labelscalefactor); dot((-15.56556976958695,-2.2144479618905266),linewidth(4pt) + dotstyle); label("$D$", (-15.467234783588566,-2.0394605133246073), NE * labelscalefactor); dot((-1.1646456874358049,-9.029282250092894),dotstyle); label("$E$", (-1.084025730582437,-8.813342327927181), NE * labelscalefactor); dot((-7.138838230212681,0.5062332301495509),linewidth(4pt) + dotstyle); label("$X$", (-7.045041727432701,0.692671818565097), NE * labelscalefactor); dot((-4.088852374348884,2.417110811275681),linewidth(4pt) + dotstyle); label("$Y$", (-3.9967949108615435,2.5893587266538174), NE * labelscalefactor); dot((-4.396583103819279,1.285229616255652),linewidth(4pt) + dotstyle); label("$Z$", (-4.312909395542997,1.4603784242200553), NE * labelscalefactor); dot((5.128824905406998,4.2699101275822295),linewidth(4pt) + dotstyle); label("$F$", (5.215684356997956,4.440886422645187), NE * labelscalefactor); dot((1.9820896089855973,-2.3796860612553314),linewidth(4pt) + dotstyle); label("$O$", (2.077119116232097,-2.1975177556653342), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $F$ be the antipode of $E$, so $AF \parallel BC$ Let $P_\infty$ denote the point at infinity along $AX$ $-1=(AX;YP_\infty)=(AE;ZF)$, $\therefore D-Z-F$, so $\measuredangle ZAD = \measuredangle AFZ = \measuredangle ZDB $, hence we're done.

Let $L=AE\cap BC$ and let $R=AZ\cap BC$. Claim 1: $\triangle BXA \sim \triangle ALD$. Proof: Since $\angle BXA = \angle ALD=90^{\circ}$, it is enough to show that $\angle XBA=\angle LAD$. However, $\angle LAD = \angle EAC+\angle CAD = \angle EBC + \angle CBA = \angle XBA$, so we are done. Now back to the problem. Note that $\angle RAD = \angle ZAD = \angle YBA$. Since $BY$ is the median in $\triangle BXA$, it follows that $AR$ is the median in $\triangle LAD.$ From PoP, we have $RZ\cdot RA = RC\cdot RB$. It is well-known that $(D,L;C,B)=-1$ and since $R$ is the midpoint of $LD,$ we obtain that $RC\cdot RB=RD^2=RZ\cdot RA$ and we are done.

Diagram. Note that $DA$ is tangent to the circumcircle of $\odot(ABC)$. Now since D lies on the diameter $BC$ and $E$ is the reflection of $A$ over $BC$, we get that $DE$ is tangent to $\odot(ABC)$ due to symmetry. Let the line parallel to $BC$ through $A$ intersect $\odot(ABC)$ again at $T$. Now note that $\measuredangle EAT = 90^{\circ}$ as $AT \parallel BC$ and $AE \perp BC$. This gives us that, \[ 90^{\circ}=\measuredangle EAT=\measuredangle EBT \implies TB\perp XE \implies XA \parallel BT \]where the last implication follows due to the fact that $XA\perp XE$. Now note that $D$ is basically the intersection of tangents to $\odot(ABC)$ at $A$ and $E$, which implies that $TD$ is the $T$-symmedian of $\triangle TAE$. Thus we have, \begin{align*} (T,TD\cap \odot(ABC);A,E)=-1&=(\infty_{XA},Y;A,X) \\ &\overset{B}{=}(B\infty_{XA}\cap \odot(ABC),A,E)\\ &= (T,Z;A,E) .\end{align*} This implies that $TD\cap \odot(ABC)\equiv Z$ which gives us that $\overline{T-Z-D}$ are collinear. Now note, \begin{align*} \measuredangle (AD, BD) = \measuredangle ADB &= \measuredangle ABD+\measuredangle DAB \\ &= \measuredangle ABC+\measuredangle ACB\\ &= \measuredangle ABC-\measuredangle BCA\\ &= \measuredangle ABC-\measuredangle TBC\\ &= \measuredangle ABT\\ &= \measuredangle AZT\\ &= \measuredangle AZD .\end{align*} This finally gives that $BD$ is tangent to $\odot(AZD)$ and we are done.

Consider $\sqrt{bc}$ inversion. New Problem Statement: $ABC$ is a triangle with $\measuredangle CAB=90$ and $AB<AC$. Let $E$ be the midpoint of $BC$ and $D$ is on $(ABC)$ such that $AD\parallel BC$. $X$ is taken on $(AEC)$ with $\measuredangle ACX=90$. Tangent to $(ABC)$ at $D$ intersects $BC$ at $T$ and $Y$ is the reflection of $A$ over $X$. Prove that $A,C,T,Y$ are concyclic. Let $A'$ be the antipode of $A$ on $(ABC)$. Note that $A',C,X$ are collinear. \[\measuredangle TEA'=2\measuredangle CAE=2\measuredangle ECA=2\measuredangle DAC=\measuredangle DEC=\measuredangle DET\]And $T$ lies on $EC$ hence $TA'$ is tangent to $(ABC)$. Since $\measuredangle XEA=90$ and $AE=EA'$, we observe that $XY=XA=XA'$ thus, $A',T,Y$ are collinear. \[180-\measuredangle ACT=\measuredangle ECA=\measuredangle CAE=\measuredangle CAA'=\measuredangle CA'T=\measuredangle XA'Y=\measuredangle A'YX=\measuredangle TYA\]Which completes the proof as desired.$\blacksquare$

Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Now we have $A'B // AX$, so $-1=(A,X;Y,\infty_{AX}) \stackrel{B}{=} (A,E;Z,A') \implies A',Z,D$ collinear. It suffices to show that $\angle BDZ = \angle DAZ$, or $\angle BDA' = \angle ABY$. Let $O$ be the centre of $\omega$, and let $P$ be the point such that $APDO$ is an isosceles trapezium with $AP//OD$. First note that $\angle OPD = \angle OAD = 90^\circ = \angle AXB$. Next, note that $PD//OE$, so $\angle ODP = \angle DOE = \frac{1}{2} \angle AOE = \angle ABX$. It follows that $\Delta ABX \sim \Delta ODP$. Now, it is not hard to see that $A'P=OD$ (by considering $A'A=2d(O,AE)$), and since $A'P//OD$, we have that $A'PDO$ is a parallelogram. So $DA'$ passes through the midpoint $M$ of $OP$. We also have $\Delta ABY \sim \Delta ODM$, so $\angle ABY = \angle ODM = \angle BDA'$ as desired. $\square$

Let $BC\cap AE=F$ and let line through $B$ parallel to $AX, CE$ meet $\odot(ABC)$ at $E$. Observe that $$(A,X;Y,P_{\infty})\stackrel{B}{=}(A,E;Z,E')\implies \overline{D-Z-E'}$$Now it is enough to show $\odot(DZFE)$ is cyclic \begin{align*} \angle DZE=\angle E'BE=\angle DFE \end{align*}as then we also have $$\angle ZDF=\angle ZEA=\angle ZE'A=\angle ZAD$$giving $\odot(ADZ)$ tangent to $BD$. [asy][asy]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 = -11.487312937104985, xmax = 2.5903843604219405, ymin = -1.6811833355429244, ymax = 9.835555034877363; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-3.304751539279967,4.132515749237636)--(-4.499987357145065,2.007950730459442)--(0.4999873571450638,1.992049269540558)--cycle, linewidth(0.6) + zzttqq); /* draw figures */ draw(circle((-2,2), 2.5), linewidth(0.6)); draw((-3.304751539279967,4.132515749237636)--(-4.499987357145065,2.007950730459442), linewidth(0.6) + blue); draw((-4.499987357145065,2.007950730459442)--(0.4999873571450638,1.992049269540558), linewidth(0.6) + blue); draw((0.4999873571450638,1.992049269540558)--(-3.304751539279967,4.132515749237636), linewidth(0.6) + blue); draw((xmin, -1.8042883891076678*xmin-6.111324209168697)--(xmax, -1.8042883891076678*xmax-6.111324209168697), linewidth(0.6) + blue); /* line */ draw((-6.765413570420523,2.015155484173907)--(-3.304751539279967,4.132515749237636), linewidth(0.6) + blue); draw((-6.765413570420523,2.015155484173907)--(-4.499987357145065,2.007950730459442), linewidth(0.6) + blue); draw(circle((-6.753116355578619,5.8818293438468805), 3.8666934140905393), linewidth(0.6) + qqwuqq); draw((-3.304751539279967,4.132515749237636)--(-3.3182891238344565,-0.12417367133193213), linewidth(0.6) + linetype("0 3 4 3") + blue); draw((-6.765413570420523,2.015155484173907)--(-0.6817108761655435,4.124173671331932), linewidth(0.6) + linetype("0 3 4 3") + blue); draw((-5.119918995728215,3.1264861879955115)--(-3.304751539279967,4.132515749237636), linewidth(0.6) + blue); draw((-6.765413570420523,2.015155484173907)--(-3.3182891238344565,-0.12417367133193213), linewidth(0.6) + blue); draw((-4.350181267644339,2.8524365133038643)--(-3.3182891238344565,-0.12417367133193213), linewidth(0.6)); draw(circle((-5.0418512892577025,0.9454907394889944), 2.028509207045663), linewidth(0.6) + qqwuqq); draw((-4.350181267644339,2.8524365133038643)--(-3.3115202147570284,2.0041710385813913), linewidth(0.6) + blue); draw((-3.304751539279967,4.132515749237636)--(-0.6817108761655435,4.124173671331932), linewidth(0.6) + blue); /* dots and labels */ dot((-2,2),linewidth(3pt) + dotstyle); label("$O$", (-2.0466181939841293,1.6071485412744468), NE * labelscalefactor); dot((-3.304751539279967,4.132515749237636),linewidth(3pt) + dotstyle); label("$A$", (-3.6377465214764086,4.228721880856959), NE * labelscalefactor); dot((-4.499987357145065,2.007950730459442),linewidth(3pt) + dotstyle); label("$B$", (-4.819727564756387,1.622302144393421), NE * labelscalefactor); dot((0.4999873571450638,1.992049269540558),linewidth(3pt) + dotstyle); label("$C$", (0.7113375736691542,1.8192989849400838), NE * labelscalefactor); dot((-3.3182891238344565,-0.12417367133193213),linewidth(3pt) + dotstyle); label("$E$", (-3.758975346428201,-0.332512657954233), NE * labelscalefactor); dot((-5.119918995728215,3.1264861879955115),linewidth(3pt) + dotstyle); label("$X$", (-5.531946911348169,3.122508853171853), NE * labelscalefactor); dot((-4.212335267504091,3.629500968616574),linewidth(3pt) + dotstyle); label("$Y$", (-4.425733883663061,3.7286529779308153), NE * labelscalefactor); dot((-4.350181267644339,2.8524365133038643),linewidth(3pt) + dotstyle); label("$Z$", (-4.698498739804594,2.910358409506216), NE * labelscalefactor); dot((-6.765413570420523,2.015155484173907),linewidth(3pt) + dotstyle); label("$D$", (-7.001846413888655,2.2132926660334094), NE * labelscalefactor); dot((-0.6817108761655435,4.124173671331932),linewidth(3pt) + dotstyle); label("$E'$", (-0.5918722945626173,4.228721880856959), NE * labelscalefactor); dot((-3.3115202147570284,2.0041710385813913),linewidth(3pt) + dotstyle); label("$F$", (-3.1982920310261598,2.152678253557513), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]