I see that nobody tryed it, so I will post my solution. Denote $\angle{MBC}=\angle{MDC}=\alpha$, $\angle{MCD}=\angle{MBA}=\beta$ and $\angle{BAC}=\angle{CAD}=t$. Thus, $\frac{MC}{\sin{\alpha}}=\frac{BC}{\sin{BMC}}$, and $\frac{MC}{\sin{\alpha}}=\frac{CD}{\sin{(\alpha+\beta)}}$ by the sin law in $\triangle{MBC}$, and $\triangle{MCD}$. So $\frac{BC}{CD}=\frac{\sin{BMC}}{\sin{(\alpha+\beta)}}$. $(1)$ Now from the sin law in $\triangle{ABC}$ and $\triangle{ADC}$, we have that: $\frac{BC}{\sin{t}}=\frac{AC}{\sin{(\alpha+\beta)}}$ and $\frac{CD}{\sin{t}}=\frac{AC}{\sin{ADC}}$ So $\frac{BC}{CD}=\frac{\sin{ADC}}{\sin{(\alpha+\beta)}}$. $(2)$ By $(1)$ and $(2)$ we have that $\sin{ADC}=\sin{BMC}$, thus $\angle{ADC}=\angle{BMC}$, or $\angle{ADC}=180-\angle{BMC}$ etc.

pohoatza wrote: I see that nobody tryed it, so I will post my solution. Denote $ \angle{MBC} = \angle{MDC} = \alpha$, $ \angle{MCD} = \angle{MBA} = \beta$ and $ \angle{BAC} = \angle{CAD} = t$. Thus, $ \frac {MC}{\sin{\alpha}} = \frac {BC}{\sin{BMC}}$, and $ \frac {MC}{\sin{\alpha}} = \frac {CD}{\sin{(\alpha + \beta)}}$ by the sin law in $ \triangle{MBC}$, and $ \triangle{MCD}$. So $ \frac {BC}{CD} = \frac {\sin{BMC}}{\sin{(\alpha + \beta)}}$. $ (1)$ Now from the sin law in $ \triangle{ABC}$ and $ \triangle{ADC}$, we have that: $ \frac {BC}{\sin{t}} = \frac {AC}{\sin{(\alpha + \beta)}}$ and $ \frac {CD}{\sin{t}} = \frac {AC}{\sin{ADC}}$ So $ \frac {BC}{CD} = \frac {\sin{ADC}}{\sin{(\alpha + \beta)}}$. $ (2)$ By $ (1)$ and $ (2)$ we have that $ \sin{ADC} = \sin{BMC}$, thus $ \angle{ADC} = \angle{BMC}$, or $ \angle{ADC} = 180 - \angle{BMC}$ etc. you did'nt prove that A,M,C are collinear ???

If $A,M,C$ are collinear ,then we have easily $\angle{BMA}=\angle{CDA}$. Assume that $A,M,C$ aren't collinear. Let $N$ $K$ be reflections of $D$ and $B$ with $MC$. Then $NBMC$ and $KMCD$ are cyclic. Since $\angle{MBA}=\angle{MCD}=\angle{MCN}$ We have: $N$ lies on the line $AB$. Let $E$ be intersection of $NB$,$DK$ and $CM$. Since $EM$ is bisector of $\angle{AED}$, $C$ is excircle center of $\triangle{EDA}$. (that touching to the side $AD$) hence $CD$ is external bisector of $\angle{ADE}$ Then: $\angle{BMC}=180^{\circ}-\angle{BNC}=180^{\circ}-\angle{CDK}=\angle{ADC}$. So we are done! (U.J.B.O.)

pohoatza wrote: Let $ABCD$ be a convex quadrilateral, with $\angle BAC=\angle DAC$ and $M$ a point inside such that $\angle MBA=\angle MCD$ and $\angle MBC=\angle MDC$. Show that the angle $\angle ADC$ is equal to $\angle BMC$ or $\angle AMB$. Remark: Actually this configuration same as Miquel's Theorem Let points P,T,S are intersection of MD,MC,MB with BC,AB,CD respectively Using above facts ,it can be shown that PBMT,TMDS,TSBC,TPCD,PBDS is cyclic And this happen if and only if T is Miquel Point of complete quadrilateral PBCDSM.

Denote $\angle BAC = \angle DAC = \alpha$, $\angle ABM = \angle MCD = x$ and $\angle MBC = \angle MDC = y$. By the Sine Law we have \[ \frac{AC}{\sin\angle ADC} = \frac{CD}{\sin \alpha} = \frac{\frac{CM}{\sin y}\sin(x+y)}{\sin \alpha} = \frac{BC}{\sin \angle BMC}\frac{\sin \angle ABC}{\sin \angle BAC} = \frac{AC}{\sin \angle BMC} \]thus $\sin \angle ADC = \sin \angle BMC$. Since both angles have measures strictly between $0^{\circ}$ and $180^{\circ}$, we either have $\angle ADC = \angle BMC$ (and we are done), or $\angle ADC + \angle BMC = 180^{\circ}$. In the latter case with $\angle BMC = z$, $\angle ADC = 180^{\circ} - z$ we obtain $\angle BCM = 180^{\circ} - y - z$ and now summing all angles in the quadrilateral $ABCD$ leads to $z = \alpha + x$. Now if $BM \cap AC = M_1$, then $\angle BM_1C = \alpha + x = z = \angle BMC$, so $M \equiv M_1$, i.e. $M$ lies on $AC$. Hence we conclude $\angle ADC = 180^{\circ} - \angle BMC = \angle AMB$, as desired.