Easily $\angle BAD= \angle BXD$. And $\angle AMX+ \angle ABX= \angle MDC= \angle DMB+ \angle DBM (\star )$ Law of sines in $\triangle AXM$: $$\frac{AX}{AM} = \frac{ \sin \angle AMX}{ \sin \angle BAD}$$ Law of sines in $\triangle ABX$: $$\frac{AB}{AX} = \frac{ \sin 90}{ \sin \angle ABX}$$ Law of sines in $\triangle ABD$: $$\frac{BD}{AB} = \frac{ \sin \angle BAD}{ \sin 90}$$ Law of sines in $\triangle BDM$: $$\frac{DM}{BD} = \frac{ \sin \angle DBM}{ \sin \angle DMB}$$ Multiplying the four equations we get: $ \frac{ \sin \angle ABX}{ \sin \angle AMX} = \frac{ \sin \angle DBM}{ \sin \angle DMB}$ $(\star \star )$ By $(\star )$ and $(\star \star )$, Two Equal Angles Lemma we get $\angle AMX= \angle DMB$. The conclusion follows.

My solution:Let XM cut (ABD) and BC at T and Y. We have:TAD=TXD=180-DXM=90. So AT parallel to BC. So ATCY is a paralellogram. So M is the midpoint of TY too.(1) Because the triangle TBY=90.(2) So MY=MB. So XMB=2MBC.

Let $N$ be the midpont of $AB$.Obviously $A,X,D,B$ are concyclic.Now, $-\measuredangle BXM=\measuredangle DXA=\measuredangle DBA=-\measuredangle BNM$. So,$B,N,X,M$ are also concyclic. Now,since $NA=NB=NX$,so,$NM$ bisects $\angle XMB$.As $MN \parallel BC$,we are done.

Let $BM$ meets $(AB)$ at $Y$ and $MX$ meets $(AB)$ at $Z$. Then we have $XY || BZ \implies XY\perp BC$ at $T$ and $MX=MY$. So we get that $\angle XMY=180^\circ-2\angle MYX=180^\circ-2\angle BNT=2\angle MBC.$

Let $Z$ be the midpoint of $AB$. $BZM= \pi - DBA= DXA = BXM$ and due to the condition on the position of X, $BZXM$ is cyclic. Now since $Z$ is the center of $(AB)$, it is the midpoint of $\overarc{BX}$ of the previously mentioned circle. Now since $MZ || BC$ we get that the bisector of $\angle XMB$ is parallel to $BC$ which completes the solution.

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[asy][asy]size(8cm); pair A=(3,6),B=(0,0),C=(10,0),D,M,Nn,X,Y; D=foot(A,B,C);M=(A+C)/2;Nn=(A+B)/2;Y=2*Nn-D;X=foot(D,M,Y); draw(circumcircle(B,D,A)^^circumcircle(D,X,M),cyan+green); D(MP("A",A,N)--MP("N",Nn,W)--MP("B",B,S)--MP("D",D,S)--MP("C",C,S)--MP("M",M,E)--A); D(M--MP("X",X,dir(275)*2)--MP("Y",Y,N),heavymagenta); draw(D--Nn--Y--A--D--M--Nn^^D--X--A^^B--X^^Y--B--M,heavymagenta); dot(A^^B^^C^^D^^X^^Y^^M^^Nn); [/asy][/asy] Let $MX$ meet $\odot (AB)$ again at $Y$, and let $N$ be the midpoint of $AB$ (i.e., the center of the circle with diameter). Then $$\angle YXD=\pi-\angle DXM=\pi-\frac{\pi}{2}=\frac{\pi}{2},$$so $Y$ is the antipode of $D$ in $\odot (AB)$, so $AYBD$ is a rectangle. Since $MN$ is clearly the perpendicular bisector of $AD$, $MY$ and $MB$ are symmetric wrt $MN$, so $$\angle XMN=\angle NMB\implies \angle XMB=\angle XMN+\angle NMB=2\angle NMB=2\angle MBC.\;\blacksquare$$

Let $N$ be the midpoint of $\overline{AB}$ and $P$ be the midpoint of $\overline{AD}$. Observe that $P$ lies on $(XDM)$ and on $\overline{MN}$. Firstly, note that $BNXM$ is cyclic quadrilateral as \begin{align*} \measuredangle NBX =\measuredangle PDX=\measuredangle PMX=\measuredangle NMX. \end{align*}Now, note that $\measuredangle NBX=\measuredangle BXN=\measuredangle BMN=\measuredangle MBC$, which means that $\overline{BX},\overline{BM}$ are isogonal wrt $\angle ABC$. Thus indeed $\measuredangle XMB=\measuredangle XMN+\measuredangle NMB=\measuredangle XBN+\measuredangle MBC=2\measuredangle MBC$. $\blacksquare$

Let the tangents at $A$ and $C$ to $(A B C)$ meet at point $K$. From $(A X D B)$ cyclic, we have $180 - \angle AXM=\angle BXD=\angle BAD$ and we can easily see that $\angle AKM=\angle BAD$, from these we get that $(A X M K)$ cyclic. Then we have $90=\angle AMK=\angle AXK$, implying that $B$, $X$ and $K$ are collinear. Since $BK$ is symmedian of $\angle ABC$, we have $\angle ABX=\angle MBD$. And we know that $90+\angle DBT=\angle BDX=90+\angle ADX=90+\angle ABX$, then $\angle DBT=\angle ABX$ (then $\angle DBT=\angle ABX=\angle MBD$). From $MX//BT$ we have $\angle XMB=\angle MBT=2\angle MBD$ which completes the solution.

Halykov wrote: X.Allaberdiyev wrote: Let the tangents at $A$ and $C$ to $(ABC)$ meet at point $K$. From $AXDB$ cyclic, we have 180 - ∠AXM=∠BXD∠=∠BAD and we can easily see that ∠AKM=∠BAD, from these we get that $AXMK$ cyclic. Then we have 90=∠AMK=∠AXK, implying that $B, X, K$ are collinear. Since $BK$ is symmedian of ∠ABC, we have ∠ABX=∠MBD. And we know that 90+∠DBT=∠BDX=90+∠ADX=90+∠ABX --> DBT=∠ABX (then ∠DBT=∠ABX=∠MBD). From $MX//BT$ we have ∠XMB=∠MBT=2∠MBD which completes the solution. Latex form of this solution

$\color{red} \textbf{Geo Marabot Solve 5}$ Let, $N$ be the mid point of $AB$ and $AD\cap MN \equiv D'.$ $\angle NMB=\angle MBC.$ Hence we need to show $\angle XMN=\angle MBC$ We have, $A,B,D,X$ concyclic and $D,D',X,M$ concyclic. So, $$\angle XMN=\angle XDD'=\angle XDA=\angle XBA=\angle XBN \implies B,N,X,M$$cyclic $$\implies \angle XBN=\angle BXN=\angle BMN=\angle MBC \blacksquare$$

Let $N$ be the midpoint of $AB$. We have that $\angle AXB = 90 = \angle ADB$, hence $A,X,D,B$ are conclycic with diameter $AB$, so $N$ is the center of $(AXDB)$, i.e. $NA=NX=ND=NB$. Also, $MN \parallel BC => MN \perp AD$ and $\angle DXM = 90 => XM \perp DX$, hence $\angle XMN = \angle XDA$ as their sides are perpendicular. From the cyclic $AXDB$ we get that $ \angle NBX = \angle ABX = \angle ADX = \angle XDA = \angle XMN => BNXM$ is cyclic and, using that $NB=NX$, we conclude that $NM$ bisects $\angle XMB$. Hence we get that $\angle XMB = 2 \angle NMB = 2 \angle MBC$, where the last equalith holds because $MN \parallel BC$.

Attachments:

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -17.24248, xmax = 19.83192, ymin = -13.56324, ymax = 8.02316; /* image dimensions */ /* draw figures */ draw((-3.2,6.98)--(-9,-5.7), linewidth(0.8) + red); draw((-9,-5.7)--(9.86,-5.52), linewidth(0.8) + red); draw((-3.2,6.98)--(9.86,-5.52), linewidth(0.8) + red); draw((-3.2,6.98)--(-3.421008752341651,-5.646754060202625), linewidth(0.8)); draw((-3.2,6.98)--(0.574,2.066), linewidth(0.8)); draw((0.574,2.066)--(-9,-5.7), linewidth(0.8)); draw((-3.421008752341651,-5.646754060202625)--(0.574,2.066), linewidth(0.8)); draw((0.574,2.066)--(3.33,0.73), linewidth(0.8)); draw((-3.421008752341651,-5.646754060202625)--(3.33,0.73), linewidth(0.8)); draw((-6.1,0.64)--(3.33,0.73), linewidth(0.8)); draw((-6.1,0.64)--(0.574,2.066), linewidth(0.8)); draw((-9,-5.7)--(3.33,0.73), linewidth(0.8)); draw(circle((-6.271508027994867,0.7184500443509642), 6.974322147214405), linewidth(0.8) + blue); draw(circle((-1.327242851760632,-5.366665643302623), 7.671977772810744), linewidth(0.8) + blue); /* dots and labels */ dot((-3.2,6.98),dotstyle); label("$A$", (-3.10968,7.22456), NE * labelscalefactor); dot((-9,-5.7),dotstyle); label("$B$", (-9.32908,-6.15804), NE * labelscalefactor); dot((9.86,-5.52),dotstyle); label("$C$", (9.95832,-5.28684), NE * labelscalefactor); dot((-3.421008752341651,-5.646754060202625),dotstyle); label("$D$", (-3.73888,-6.23064), NE * labelscalefactor); dot((3.33,0.73),linewidth(4pt) + dotstyle); label("$M$", (3.42432,0.93256), NE * labelscalefactor); dot((0.574,2.066),dotstyle); label("$X$", (0.66552,2.31196), NE * labelscalefactor); dot((-6.1,0.64),linewidth(4pt) + dotstyle); label("$N$", (-6.64288,0.69056), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\color{red}\textbf{Claim:-}$ $\angle XMB=2 \angle MBC.$ $\color{blue}\textbf{Proof:-}$ Let the midpoint of $AB$ is $N.$ From this we get, $A,B,D,X$ are cyclic, Since, $AD\perp BC\implies\angle AXB=\angle ADB=90^o.$ Now Since, $MN\parallel BC\implies \angle BMN=\angle MBC$ also we get, $\angle BNM+\angle ABC=180^o\implies \boxed{\angle BNM=180^o-\angle ABC.}$ Now We claim that $MN$ bisects $\angle XMB.$ Since, $A,B,D,X$ are cyclic we get, $\angle BXD=\angle DAB=90^o-\angle ABC.$ Next we claim that the circle through $A,B,D,X$ has center at $N$ and $AB$ is diameter. Its very obvious since, $AB$ subtends $90^o$ at $D$ and $X\implies AN=NX=BN.$ Now in triangle $BNX\implies NX=NB\implies\angle NBX=\angle NXB.$ Now we also see that $B,N,X,M$ are cyclic. Since, $\angle BNM=\angle BXM=180^o-\angle ABC.$ Now in Cyclic Quad. $B,M,X,N$ we get, $\angle NXB=\angle NMX\implies \boxed{\angle XMB=2\angle MBC.}$