Let $M$ be the midpoint of $\overline{AB}$ and $\ell$ the perpendicular bisector of $\overline{AD}$. Then $OM,OZ$ are symmetric in $\ell$, so $OM$ passes through $Z'$, the reflection of $Z$ in $\ell$. Now $AZZ'D$ is a rectangle and $\angle AMZ'=90^{\circ}$, so we're done.

Apply inversion at $A$. We obtain the equivalent problem: inverted APMO 2017/2 wrote: In triangle $ABC$ with $AB>AC$, let $M$ be a point on $AB$ such that $BA=BM$ and $M \ne A$, $D$ be the foot of the $A$ angle bisector, $Z$ be a point on the external bisector of angle $A$ such that $CA=CZ$. Prove that $M, D, Z$ are collinear. (Proof) To see this, let $X=MD \cap AC$ and $Y=MZ \cap AC$. Note that $AB \parallel CZ$ and $CZ=CA$ so $$\frac{AY}{YC}=\frac{AM}{CZ}=\frac{2AB}{AC}.$$Finally, by Menelaus' Theorem in $\triangle ABC$ for transversal $MDX,$ $$\frac{AX}{XC} \cdot \frac{CD}{DB} \cdot \frac{BM}{MA}=1 \Longrightarrow \frac{AX}{XC}=\frac{2AB}{AC}.$$The last result is due to the Angle-Bisector Theorem. Evidently, $X \equiv Y$ so we are done! $\blacksquare$

Let $(AZD)\cap AB =K$ .By angle chasing we get triangles $ZKD$ and $ZNC$ are similar.So there exist a spiral similarity centered at Z that takes KD to NC .Then by lemma if $KN\cap CD=O$ $OZKD$ is cyclic.Again angle chasing gives $KN\parallel BC$.So done!!!(here N is midpoint of AC).

My solution in the contest : Let $O$ the circumcenter of $\odot(ABC)$, $M$ the midpoint of $AB$, let $E$ the midpoint of the arc $\overarc {BAC}$ and $N$ the midpoint of $EZ$, so from $AEDB$ is cyclic we get: $\measuredangle ABD=\measuredangle ZED=\measuredangle ZEO...(1)$, so from $DA\perp AZ$ we get $\measuredangle BAD$ $=$ $\measuredangle DAC$ $=$ $\measuredangle AZO$ $=$ $\measuredangle EZO..(2)$ Combining $(1)$ and $(2)$ we get $\triangle BAD\cup \{M\}\sim \triangle EOZ\cup \{N\}$ $\Longrightarrow$ $\measuredangle BMD=\measuredangle ENO$, so from $NO\parallel ZD$ we get $\measuredangle BMD$ $=$ $\measuredangle ENO$ $=$ $\measuredangle EZD=\measuredangle AZD$ hence $AZDM$ is cyclic.

Solution 1: Define $\{A,M\}\equiv \odot(ADZ)\cap AB, \{A,X\}\equiv \odot(ADZ)\cap AC$. Notice that $D$ is the spiral center sending segment $\overline{CX}$ to $\overline{BM}$, and since $DB=DC$, this implies that $BM=CX$. It is also easy to check that $BA$ is tangent to $\odot(AZC)$, which quickly implies that $Z$ is the spiral center sending segment $\overline{AM}$ to $\overline{CX}$, and since $ZA=ZC$, this implies that $AM=CX$. Therefore, $AM=BM$, and the conclusion follows. Solution 2: Invert about $A$ with power $r^2=\tfrac{1}{2}AB\cdot AC$, composed with a reflection about the $A$-angle bisector. The problem now becomes to show that if $Z$ is the projection of $C$ onto the external angle bisector of $\angle BAC$, and if the internal angle bisector of $\angle BAC$ intersects $\overline{BC}$ at $X$, then $BZ$ bisects $\overline{AX}$ at a point $D$. It is well-known that $Z$ is collinear with the midpoints $M_B, M_A$ of $\overline{AC}, \overline{BC}$, respectively, so we have $$\frac{ZM_B}{ZM_A}=\frac{M_BC}{M_BC+M_BM_A}=\frac{AC}{AC+AB}=\frac{\tfrac{BC\cdot AC}{2(AC+AB)}}{\tfrac{BC}{2}}=\frac{DM_B}{BM_A}$$so $\triangle ZDM_B\sim\triangle ZBM_A$, as desired.

After I finished this, I realized that complex would have probably been faster, but this was only 10 minutes so I didn't mind

Let $A'$ be the reflection of $A$ across $Z$ and $B'$ be the reflection of $B$ across $D$. It is easy to see that $CB'B$ and $CAA'$ are similar right triangles so $AB'\perp A'B$, which means that $\angle DMZ=90^\circ$ as $AB'\parallel DM,A'B\parallel MZ$, immediately implying the conclusion.

Did I mess up in not explicitly incorporating $AB < AC$?

Redefine $Z$ to be the point on $(AMD)$ such that $\overline{AZ}$ is an external bisector; hence $\measuredangle DMZ = \measuredangle DAZ = 90^{\circ}$. Then let $N$ be the midpoint of $\overline{AC}$. As $\measuredangle AMN = \measuredangle ABC = \measuredangle ADC$, we find $T = \overline{MN} \cap \overline{CD}$ lies on $(AMD)$. [asy][asy] pair A = dir(110); pair B = dir(200); pair C = dir(340); draw(unitcircle, lightred); draw(A--B--C--cycle, lightred); pair D = dir(270); pair M = midpoint(A--B); pair Z = extension(origin, A+C, A, dir(90)); pair N = midpoint(A--C); draw(circumcircle(A, D, Z), dashed+heavycyan); draw(A--Z, lightred); draw(A--D, lightred); draw(D--M--Z--cycle, heavycyan); pair T = extension(M, N, D, C); draw(M--T--D, dashed+cyan); pair L = dir(90); draw(circumcircle(Z, T, C), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$N$", N, dir(225)); dot("$T$", T, dir(T)); dot("$L$", L, dir(L)); [/asy][/asy] Let $L$ be the $D$-antipode. We have $\triangle ZMD \sim \triangle LBD$ also by spiral similarity, so \[ \frac{ZM}{ZD} = \frac{LB}{LD} = \cos A/2 = \frac{MN}{DC} \]plus $\measuredangle ZMN = \measuredangle ZMT = \measuredangle ZDT = \measuredangle ZDC$. So we obtain the direct similarity $\triangle ZMN \sim \triangle ZDC$, hence $\triangle ZMD \sim \triangle ZNC$, so $\overline{ZN} \perp \overline{AC}$.

Man, my oly geo is pretty rusty Let $M$, $N$, and $E$ be the midpoints of $\overline{AB}$, $\overline{AC}$, and $\overline{BC}$ respectively, noting that $DE\perp BC$. Remark that $\triangle CED\sim\triangle ZNC$, so \[\dfrac{DE}{EM}=\dfrac{DE}{CN}=\dfrac{DC}{CZ}.\]Furthermore, \[\angle DEM=90^\circ+\angle MEB = 90^\circ + \angle ACB = (\angle ZCN+\angle DCE)+\angle ACB=\angle ZCD.\]Hence $\triangle DEM\sim\triangle DCZ$, so by spiral similarity \[\triangle DEC\sim\triangle DMZ\quad\implies\quad \angle DMZ=\angle DEC = 90^\circ.\]Combining this with $\angle DAZ=90^\circ$ establishes the desired cyclicity. [asy][asy] size(250); defaultpen(linewidth(0.8)); pair A = dir(120), B = dir(215), C = dir(325), D = dir(270), M = (A+B)/2, N = (A+C)/2, E = (B+C)/2, P = dir(90), Z = extension(A,P,origin,N); draw(A--B--C--A--Z--C^^Z--N^^B--D--C^^unitcircle); draw(circumcircle(D,A,Z),linetype("4 4")); draw(D--M--E--cycle,linetype("8 8")); label("$A$",A,dir(origin--A)); label("$B$",B,dir(origin--B)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); dot(M^^N^^E); label("$M$",M,dir(origin--M)); label("$Z$",Z,dir(N--Z)); label("$E$",E,NE); label("$N$",N,dir(Z--N)); draw(rightanglemark(Z,N,C,3)^^rightanglemark(D,E,C,3)); [/asy][/asy]

Dear Mathlinkers, an outline of my proof 1. the circle with diameter CZ intersect (AZD) in T 2. TN is parallel to BC (angle chasing) 3. Reim's theorem for (O) and (AZD) and we are done... Sincerely Jean-Louis

Took me 20 minutes during contest

ABCDE wrote: Let $A'$ be the reflection of $A$ across $Z$ and $B'$ be the reflection of $B$ across $D$. It is easy to see that $CB'B$ and $CAA'$ are similar right triangles so $AB'\perp A'B$, which means that $\angle DMZ=90^\circ$ as $AB'\parallel DM,A'B\parallel MZ$, immediately implying the conclusion. Could you please explain more, why is $AB'\perp A'B$? Thank you!

$CAB'$ and $CA'B$ are spirally similar with an angle of $90^\circ$.

Let $M $ and $N $ be the midpoints of $AB $ and $AC $ respectively. We have $\angle MNZ = \angle DCZ = 90^{\circ} + \angle ACB $. Observe that $MN = \frac {BC}{2}$. So, $\frac {ZN }{ZC} = \frac{MN}{CD} = \cos \frac {\angle BAC}{2} $. So, $Z $ is the center of the spiral similarity carrying $MN\rightarrow DC$. Thus, $Z $ is also the center of the spiral similarity taking $NC\rightarrow MD $. So, $\angle DMZ $ = $90$°.

I think the following fast solution is not appeared above Let $P$ be the reflection of $D$ through $M$. Then, $AP=BD=DC$, $CZ=AZ$ and $\angle{PAZ}=\angle{DCZ}=90^{\circ}+\angle{ACB}$. Therefore the triangles $PAZ$ and $DCZ$ are equal, so $DZ=PZ$ and thus $ZM\perp MD$.

Hi . Long time no see.

Let $M $ and $N$ be midpoints of $AB$ and $ AC $, $ CD \cap MN = {E}$ $\angle MNZ = 90^{\circ}+\angle C = \angle DCZ$ So $NZEC$ cyclic but $\angle ZNC = 90^{\circ} \Rightarrow \angle ZEC=90^{\circ}$ $\Rightarrow$ $ZEDA$ cyclic $\rightarrow (1)$ $MN // BC$ $\Rightarrow \angle MED = \angle BCD = \angle BAD = \angle MAD$ $\Rightarrow MAZE $ cyclic $\rightarrow (2)$ $(1),(2)$ $\Rightarrow (ADZ) $ passes through $M$

This solution was devised by my friend, Thanic Nur Samin, who isn't posting because he has some silly personal secret to protect. Sadly I made no progress on this relatively easy problem, or else I'd finally get that APMO medal We use cartesian coordinates. Let the external biscector be the $x$-axis and the internal biscector be $y$ axis. So $A$ is the origin. Let $t$ be the slope of $AC$. So the slope of $AB$ is $-t$. So we can set $B\equiv (2q,-2qt)$ and $C\equiv (2p,2pt)$. So $M\equiv (q,-qt)$. Set $D\equiv (0,d)$ and $Z\equiv (z,0)$. Now, note that, $$ZA=ZC$$$$\Rightarrow z^2=(z-2p)^2+(2pt)^2$$$$\Rightarrow z^2-(z-2p)^2=(2pt)^2$$$$\Rightarrow 2p(2z-2p)=(2pt)^2$$$$\Rightarrow z=p(t^2+1)$$ Also, note that, since $\angle DAB=\angle CAD$ and $ABCD$ is cyclic, $DB=DC$ $$\Rightarrow (2q)^2+(d+2qt)^2=(2p)^2+(d-2pt)^2$$$$\Rightarrow (d+2qt)^2-(d-2pt)^2=(2p)^2-(2q)^2$$$$\Rightarrow (2d+2qt-2pt)(2qt+2pt)=4(p+q)(p-q)$$$$\Rightarrow (d+qt-pt)t=p-q$$$$\Rightarrow dt-(p-q)t^2=p-q$$$$\Rightarrow dt=(p-q)(t^2+1)$$$$\Rightarrow d=\dfrac{(p-q)(t^2+1)}{t}$$[We have to cancel out $(p+q)$ from both sides in a step, which is nonzero since $AB<AC$] Now, let $m=p/t$ and $n=q/t$. So $p=mt$ and $q=nt$. Therefore $D\equiv (0,(m-n)(t^2+1))$ $M\equiv (nt,-nt^2)$ $Z\equiv (mt(t^2+1),0)$ So, the slope of $MD$ is $\dfrac{\text{rise}}{\text{run}}=\dfrac{(m-n)(t^2+1)+nt^2}{-nt}=\dfrac{m(t^2+1)-n}{-nt}$ And the slope of $MZ$ is $\dfrac{\text{rise}}{\text{run}}=\dfrac{nt^2}{mt(t^2+1)-nt}=\dfrac{nt}{m(t^2+1)-n}$ Since their product is $-1$, we can conclude that $MD$ and $MZ$ are perpendicular, so $\angle DMZ=90^{\circ}$. Since $\angle DAZ=90^{\circ}$, $M$ lies on the circumcircle of $\triangle ADZ$ and we are done.

Let $M$ and $N$ be the midpoints of $AB$ and $AC$ and let $X = MN \cap CD$. Since $MN$ and $BC$ are parallel, $\angle MXD = \angle BCD = \angle MAD$ so the points $A,M,D,X$ are concyclic. Also, $\angle NZC = \angle NAC$ so $Z,N,C,X$ are concyclic too. To finish, note that $\angle ZXD = \angle ZXC = 90 = \angle ZAD$, so $AMDZ$ is cyclic, as desired. $\blacksquare$

Suppose the midpoint of $\overline{AB}$ is $M$, and the midpoint of $\overline{BC}$ is $N$. I claim that $\triangle DCZ \sim \triangle MNZ$. Note that $\angle MNZ = \angle MNA + \angle ANZ = C + 90^{\circ}$. Also, $\angle DCZ = \angle DCB + \angle BCA + \angle ACZ = \frac{A}{2} + C + 90^{\circ} - \frac{A}{2} = C + 90^{\circ}$. Now, note that $\frac{NZ}{ZC} = \sin \left(90^{\circ} - \frac{A}{2} \right) = \cos \left( \frac{A}{2} \right) $ and $\frac{MN}{DC} = \frac{\frac{a}{2}}{\frac{a \cos \frac{A}{2}}{2}} = \cos \left(\frac{A}{2} \right)$, which implies that $\frac{NZ}{ZC} = \frac{MN}{DC}$, which implies that $\triangle DCZ \sim \triangle MNZ$ by ASA. Thus, $\angle DZC = \angle MZN$, which implies that $\angle MZD = \angle NZC = 90^{\circ} - \left(90^{\circ} - \frac{A}{2} \right) = \frac{A}{2}$, which implies that $\angle MZD = \angle MAD$, which implies that $M, A, Z, D$ are concyclic.

Quite simple. Taking $\sqrt{BC}$ through $A$ gives $BZ' || AC$ and $\overline {Z'AZ}$, $D' = AD \cap BC$, $M'$ is the reflection of $A$ over $C$. It suffices to show $\overline {Z'D'M'}$ with $\frac{BD'}{D'C} = \frac{Z'B}{CM'}$. However if $B'$ is the reflection of $B$ over $AD$ then since $AZ'BB'$ is a parallelogram, $BZ' = AB' = AB$ and $CM' = AC$ so this follows from angle bisector theorem.

Let $E$ be the midpoint of $AC$, and define $M$ to be the intersection of $(ZAD)$ with line $AB$ Claim: $Z$ is the spiral center sending $CE$ to $MD$ Proof: Note that $\angle{ZMD}=\angle{ZAD}=90=\angle{ZEC}$ $\angle{CZE}=\angle{EZA}=\angle{CAD}=\angle{DAM}=\angle{DZM}$ So, $\triangle{CZE} \sim \triangle{DZM}$ Using our claim, we have that $Z$ is the spiral center sending $CD$ to $EM$ Which implies $\angle{ZEM}=\angle{ZCD}=\angle{ZCA}+\angle{ACB}+\angle{BCD}=90+\angle{ACB}$ $90+\angle{ACB}=\angle{ZEM}=90+\angle{AEM}$ This implies that $EM$ parallel to $CB$, since $E$ is the midpoint of $AC$ it means that $M$ must be the midpoint of $AB$

this is pretty contrived but whatever Let $E$ be the second intersection of $AZ$ with $(ABC)$ and $M$ be the midpoint of $\overline{AB}$. Since $\overline{EA} \perp \overline{AD}$, $E$ must be the midpoint of arc $BAC$ of $(ABC)$. A quick angle chase now gives $\angle BEC = \angle BAC = \angle AZC$, and since $\triangle EBC$ and $\triangle ZAC$ are both isosceles, there must a spiral similarity at $C$ taking $\triangle EBC$ to $\triangle ZAC$. In particular, this spiral similarity takes $\overline{BE}$ to $\overline{AZ}$, so there also exists a spiral similarity $\tau$ at $C$ taking $\overline{BA}$ to $\overline{EZ}$. So, $\tau(M) = M'$ is just the midpoint of $\overline{EZ}$. Claim: $\tau(D)$ is the center of $(ABC)$. Proof: Let $O$ be the center of $(ABC)$. Then $O$ is the midpoint of diameter $\overline{ED}$, and $O$ lies on the perpendicular bisector of $\overline{EC}$. However, since $\angle DEC = \angle DBC$ and $\triangle ABC \sim \triangle ZEC$ (so lines $BD$ and $ED$ "correspond" to each other), we find that $\tau(D)$ must lie on $\overline{ED}$. But $\tau(D)$ is also the intersection of the internal bisector of $\angle EZC$ with $(EZC)$ and hence lies on the perpendicular bisector of $\overline{EC}$, so indeed $\tau(D) = O$. To finish, since $\overline{OM'} \parallel \overline{DZ}$, we get $$\measuredangle AMD = \measuredangle ZM'O = \measuredangle M'ZD = \measuredangle AZD,$$so $M$ indeed lies on $(AZD)$.

Let $M$ be the midpoint of $AB$ and $E$, $F$ be the midpoints of arcs $\overarc{CA}$, $\overarc{AB}$. $\textcolor{blue}{\textbf{Claim 1:}}$ The perpendicular bisector of $AD$ is parallel to $EF$. \begin{align*} \angle (AC, EF) &= \frac 12 \left(\overarc{AF} + \overarc{DE}\right) \\ &= \frac 12 \left(\angle A + \angle B + \angle C\right) \\ &= 90.~{\color{blue} \Box} \end{align*} $\textcolor{blue}{\textbf{Claim 2:}}$ $OE$ and $OF$ are symmetric about the perpendicular bisector of $AD$. First we note $O$ lies on the perpendicular bisector. Since $\triangle OEF$ is isosceles, $\angle OEF = \angle OFE$, and we have the conclusion by alternate interior angles. ${\color{blue} \Box}$ $\textcolor{blue}{\textbf{Claim 3:}}$ Pentagon $MYZAD$ is cyclic. We know $AD$ and $AZ$ are the internal and external bisectors of $\angle BAC$, respectively, so $\angle ZAD = 90$. As a result, we can see each half of $YZAD$ is a rectangle, so $YZAD$ itself is a (cyclic) rectangle with diameter $AY$. We also have \[\angle AMY = \angle AMO = 90,\] so $M$ also lies on $(ADZ)$. $\blacksquare$ [asy][asy] size(250); pair A, B, C, O, M, X, D, E, F, Z, Y; A = dir(120); B = dir(210); C = dir(330); O = (0, 0); M = .5A + .5B; X = dir(90); D = dir(270); E = dir(45); F = dir(165); Z = extension(A, X, O, E); Y = D + Z - A; draw(A--B--C--cycle^^circumcircle(A, B, C)); draw(Y--F^^O--Z); draw(D--A--Z--Y--cycle^^E--F); draw(dir(195)--foot(O, Y, Z), red); draw(circumcircle(A, Y, Z), blue+dashed); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, 2dir(0)); dot("$F$", F, W); dot("$M$", M, dir(210)); dot("$Y$", Y, SE); dot("$Z$", Z, NE); dot("$O$", O, S); [/asy][/asy]

Apply $\sqrt{bc}$ inversion at $A$, which yields the following problem. Restated problem wrote: Let $ABC$ be a triangle. Let $E$ be the intersection of the internal $\angle A$-bisector and $\overline{BC}$, $A'$ be the reflection of $A$ over $C$, and $X$ be the point on the external $\angle A$-bisector such that $\overline{BX} \parallel \overline{AC}$. Prove that $X,E,A'$ collinear. The reason this is the inverted problem statement is because $Z$ gets sent to the point $X$ on the external $\angle A$-bisector such that $\measuredangle ACZ=\measuredangle AXB$, so $\triangle ACZ \sim \triangle AXB$. But since $\angle ZAC=90^\circ-\angle A/2$, $\angle ABX=\angle AZC=\angle A$, hence $\overline{BX} \parallel \overline{AC}$. It then suffices to prove that $\frac{EB}{EX}=\frac{EC}{EA'} \iff \frac{EB}{AB}=\frac{EC}{AC}$. This is just the angle bisector theorem. $\blacksquare$

We barybash. Set $\triangle ABC$ as reference and note that it suffices to show that $Z \in (AMD)$. Note that $D$ can be parametrized as $(t: b: c)$. Plugging this into circle equation we have, \begin{align*} a^2bc + b^2tc + c^2tb &= 0\\ \iff t &= -\frac{a^2bc}{b^2c + bc^2}\\ \iff t&= -\frac{a^2}{b + c} \end{align*}So we have $D = (-a^2 : b(b+c) : c(b+c))$. Now we move onto computing $Z$. Note that it can be parametrized as $(t: -b : c)$. This should satisfy the equation of the perpendicular bisector of $\overline{AC}$ given by, \begin{align*} 0 = b^2(z - x) + y(c^2 - a^2) \end{align*}Plugging in we have, \begin{align*} 0 &= b^2(c - t) + (-b)(c^2 - a^2)\\ \frac{(c^2 - a^2)}{b} &=c - t\\ t &= \frac{bc - c^2 + a^2}{b} \end{align*}Thus we have $Z = (a^2 + bc - c^2, -b^2, bc)$. Now consider the equation of the circle $(AMD)$. It is given for constants $v$ and $w$ by, \begin{align*} -a^2yz - b^2xz - c^2xy + (vy + wz)(x+y+z) = 0 \end{align*}Plugging in the coordinates of $M = (1 : 1 : 0)$ we find that, \begin{align*} -c^2 + 2v &= 0\\ v &= \frac{c^2}{2} \end{align*}Then plugging in the coordinates of $D$ we find that, \begin{align*} -a^2bc(b+c)^2 + a^2b^2c(b+c) + a^2bc^2(b+c) + \left(\frac{bc^2(b+c)}{2} + wc(b+c) \right)(-a^2 + (b+c)^2) &= 0\\ -a^2bc(b+c) + a^2b^2c + a^2bc^2 + \left( \frac{bc^2}{2} + wc \right)(b + c - a)(a + b + c) &= 0\\ \left(\frac{bc^2}{2} + wc \right)(b + c - a)(a + b +c) &= 0\\ \frac{bc^2}{2} + wc = 0 \end{align*}so we find $w = \frac{-bc}{2}$. Then our circle has equation, \begin{align*} -a^2yz - b^2xz - c^2xy + \left(\frac{c^2}{2}y - \frac{bc}{2}z\right)(x+y+z) = 0 \end{align*}It is easy to verify $Z$ satisfies this so we're done.

We $\sqrt{bc}$ invert. The new problem becomes: Quote: In triangle $\triangle ABC$, $M$ is the reflection of $A$ across $B$, and $D$ is the foot of the internal bisector of $\angle A$ to $BC$, and $Z$ is the point on the external bisector such that $AC=ZC$. Prove that $M,D,Z$ are collinear. Let $T$ be the intersection of $AZ$ and $BC$. We will Menalaus on $\triangle ABT$. We have by ratio lemma that $$\frac{ZA}{ZT}=\frac{\sin\alpha}{\sin\beta}\cdot\frac{AC}{CT}=\frac{\sin\alpha}{\sin\beta}\cdot\frac{\sin\angle ATC}{\sin90-\alpha/2},$$$$\frac{MB}{MA}=\frac{1}{2},$$and $$\frac{DT}{DB}=\frac{1}{\sin\alpha/2}\cdot\frac{AT}{AB}=\frac{1}{\sin\alpha/2}\cdot\frac{\sin\beta}{\sin \angle ATC}.$$ Thus, $$\frac{ZA}{ZT}\cdot\frac{MB}{MA}\cdot\frac{DT}{DB}=\frac{\sin\alpha}{2\sin(90-\alpha/2)\sin\alpha/2}=1,$$as desired.

Let $K$ denote the midarc of $BAC$, and define $Z$ as $(AMD)\cup AK$, let $T$ be $MN \cup DC$, $\angle DTM =\angle DCB=\angle DBC=\angle MAD$, thus $T$ lies on $(AMD)$, by spiral similarity we have that $\triangle ZMD ~ \triangle KBD$, thus $\frac{ZM}{ZD}=\frac{KB}{KD}$, let $R$ denote $KD\cup BC$, we have that $RK$ is the perpendicular bisector of $BC$, and thus $MN=BR$, we also have that $\angle KBD=90$ so $\triangle KBR ~ \triangle KBD$, so $\frac{KB}{KD}=\frac{BR}{BD}=\frac{MN}{BD}$, finally we have $\angle ZMN=\angle ZDC$, thus $\triangle ZMN ~ \triangle ZDC$, so by spiral similarity $\triangle ZNC ~ \triangle ZMD$ which suffices.

Used Evan's diagram for this problem, but with a different solution: Let $D$ be the midpoint of arc $BC$ not containing $A,$ let $M$ be the intersection of $(ADZ)$ and $AB,$ and let $N$ be the midpoint of $AC.$ Observe $\angle ZMD=\angle ZAD=90^\circ$ and $\angle ZNC=90^\circ.$ Next, \[\angle CZN=\angle ZAN=\angle CAD=\angle MAD=\angle MZD.\]Thus $\triangle ZMD \sim \triangle ZNC.$ Then there exists a unique point $T=MN \cap CD$ that is on $(ZMD)$ and $(ZNC)$ by the fundamental lemma of spiral similarity. Now, \[\angle DTM=\angle DAM=\angle DAB=\angle DCB\]so $TM \parallel BC$ and $M$ must be the midpoint of $AB.$

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Now, invert at $A$ and refer to the images of points (other than $A$) by their original names. Let $Y$ be the foot from $M$ to the external bisector of $A$ so that $YMNZ$ is a trapezoid. Let $D'$ be the reflection of $A$ across $D$. It's a well known property of trapezoids that, since $YA:AZ = YM : ZN$ and $D$ is the midpoint of $AD'$, $D$ is the intersection of the diagonals $MZ$ and $YN$. In particular, $M$, $D$ and $Z$ are collinear, as desireed.

Perform $\sqrt{\frac{bc}{2}}$-inversion, we have - $E \rightarrow C$ ($E$ is midpoint of $AB$), $H \rightarrow B$ - $D \in (ABC) \implies D \rightarrow I$ where $I$ is intersection of $AF$ and $EH$ - $G$ goes to feet of perpendicular from $B$ to $AZ$ - $A \rightarrow A$ Suffice to prove that $G,I,C$ collinear, as it is equivalent to $(AEDZ)$ cyclic. Let $K = AC \cap GB$, and redefine $G$ to be the intersection of parallel to $AF$ through $B$ and $CJ$, we wish to prove that $AG \perp BG$. Indeed, $\frac{AI}{IF} = \frac{KG}{GB}$ so $G$ is midpoint of $KB$ and $\angle BKC = \angle FAC = \angle BAF = \angle KBA$ so we are done. $\blacksquare$

I don't have inversion skills yet, so lets prove classically. Let $J$ and $H$ be midpoints of sides $AB$ and $AC$, respectively. Let $JH$ meet $DC$ at $K$. Since $\angle AJH=\angle ABC=\angle ADC$, we have points $A$, $D$, $K$ and $J$ are concyclic. On the other hand by using the concyclic property we reached $\angle BAD=\angle HKC=\angle HZC\angle DAC=\angle HZC$ which implies points $C$, $H$, $K$ and $Z$ are concylic. We complete the proof as $$\angle ZKC=\pi-\angle ZHC=90^{\circ}=\angle DBZ$$implying $Z\in (AJDK)$ as desired.

spiral spiral yes yes Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Note that the problem statement is true iff $\triangle ZMD \sim \triangle ZNC$, a spiral similarity centered at $Z$. Thus we complete the spiral similarity by considering $MN \cap CD = K$. $ZNCK$ should be cyclic, so we let $\odot ZNC \cap MN = K'$ and prove $K=K'$. We have \begin{align*} \angle NZK' &= 180 - \angle ZNK' - \angle ZKN \\ &= 180 - (90 - \angle C + \angle ZCN) \\ &= 90 + \angle C - \angle CAZ \\ &= \angle C +\frac{\angle A}{2} \\ \angle NZK' &= \angle ACD. \end{align*}Thus $DCK'$ is collinear and $K=K'$. Then $AZKD$ is cyclic since $\angle ZKD = 90$, so it is sufficient to prove that $AMZK$ is cyclic. This is evident since \[ \angle ZKM = \angle ZCN = \angle ZAN = 180 - \angle MAZ. \]$\blacksquare$