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Let A' = DM^BC, from <CMD = <MCB+<MAB it follows that ABA'M is ciclyc. Let C' = BM^AD, from <AMB = <MAD+<MCD it follows that DMCC' is ciclyc. As supplementary of the same angle, it holds that <DMA = <CBA. From collinearity problem we have that <ACB = <DAM. Then DM/AM = AB/BC and finally DM/CM=AB/BC. As both supplementary of the equal angles <C'MC = <C'DC, it holds <BMC = <ADC. Furthermore, see again collinearity problem, <MCB = <MCA+<ACB = <MAC+<DAM = <DAC. Then BM/MC=CD/AD or BM/MA=CD/AD.

OFFICIAL SOLUTION: Construct the convex quadrilateral $PQRS$ and the interior point $T$ such that $\triangle PTQ \equiv \triangle AMB, \triangle QTR \sim \triangle AMD$ and $\triangle PTS \sim \triangle CMD$. It follows that \[ TS = \frac{MD \cdot PT}{MC} = MD, \hspace{0,2cm} \frac{TR}{TS} = \frac{MD \cdot MB}{MA \cdot MD} = \frac{MB}{MC} \] and $\widehat{STR} \equiv \widehat{BMC}$, therefore $\triangle RTS \sim \triangle BMC$. The assumption on angles leads to \[ \widehat{QPS} + \widehat{RSP} = \widehat{QPT} + \widehat{TPS} + \widehat{TSP} +\widehat{TSR} = \widehat{PTS} + \widehat{TPS} + \widehat{TSP} = 180^{\circ} \] and \[ \widehat{RQP} + \widehat{SPQ} = \widehat{RQT} + \widehat{TQP} + \widehat{TPQ} +\widehat{TRS} = \widehat{QTP} + \widehat{TQP} + \widehat{TPQ} = 180^{\circ}, \] so $PQRS$ is a parallelogram. Hence $PQ = RS$ and $QR = PS$, that is \[ AB = \frac{BC \cdot TS}{MC} = \frac{BC \cdot MD}{MC} \hspace{0,2cm} and \hspace{0,2cm} \frac{AD \cdot QT}{AM} = \frac{CD \cdot TS}{MD} = CD. \] The conclusion is now obvious.

orl wrote: The point $ M$ is inside the convex quadrilateral $ ABCD$, such that \[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \] Prove that $ AB \cdot CM = BC \cdot MD$ and $ BM \cdot AD = MA \cdot CD.$

Let $l$ be the perpindicular bisector of $AC$. We note that the exterior angle of quadrilateral $AMCD$ at $M$ is equal to the sum of the interior angles at $A,C,D$, which leads to $\angle BMC=\angle ADC$. Similarly $\angle AMD=\angle ABC$. Now construct $B'$ so that $\triangle AB'C \sim \triangle AMD$. Then $\angle AB'C=\angle AMD=\angle ABC$. But the construction will also make $\triangle ADC \sim \triangle AMB'$, so $\angle AMB' =\angle ADC=\angle BMC$. Since $l$ bisects $\angle AMC$, it bisects $BMB'$ as well. So $B'$ must be the reflection of $B$ in $l$, which makes $\triangle CBA \simeq \triangle AB'C \sim \triangle AMD$, and the desired result quickly follows.

Hint: $ACB \equiv AMD$.

I have solved it in a different way, I have used an inversion with center $M$. Here is my solution : Consider an inversion with center $M$ and radius $r=MA=MC$. Let $X'$ be the image of $X$ under that inversion. We know that $A'=A$ and $C'=C$. Since $\angle{AMB} = \angle{MAD} + \angle{MCD}$, So $\angle{A'MB'} = \angle{MD'A'} + \angle{MD'C'} \Longrightarrow \angle{A'MB'} = \angle{A'D'C'}$ And since $\angle{CMD} = \angle{MCB} + \angle{MAB}$, So $\angle{C'MD'} = \angle{MB'C'} + \angle{MB'A'} \Longrightarrow \angle{C'MD'} = \angle{A'B'C'}$ We need to prove that: $$AB \cdot CM=BC \cdot MD \Longleftrightarrow \frac{r^2 A'B'} {MA' \cdot MB'}\cdot \frac{r^2}{MC'}=\frac{r^2 B'C'} {MC' \cdot MB'}\cdot \frac{r^2}{MD'}$$$$\Longleftrightarrow \frac{A'B'}{B'C'} =\frac{MA'}{MD'} \Longleftrightarrow \frac{A'B'}{B'C'} =\frac{MC'}{MD'}$$We also need to prove that : $$BM \cdot AD=MA \cdot CD \Longleftrightarrow \frac{r^2 A'D'} {MA' \cdot MD'}\cdot \frac{r^2}{MB'}=\frac{r^2 D'C'} {MC' \cdot MD'}\cdot \frac{r^2}{MA'}$$$$\Longleftrightarrow \frac{A'D'}{D'C'} =\frac{MB'}{MC'} \Longleftrightarrow \frac{A'D'}{D'C'} =\frac{MB'}{MA'}$$So the problem is equivalent to this problem: The point $M$ is inside the quadrilateral $ABCD$ such that $MA=MC, \angle{CMD} =\angle{ABC} \quad \textnormal{and} \quad \angle{AMB} =\angle{ADC} $. Prove that $\frac{AD}{DC} =\frac{MB}{MA}$ and $\frac{AB}{BC} =\frac{MC}{MD}$ Let $T$ be a point such that $\triangle ABM \sim \triangle ACT$ ($ADTC$ is a cyclic quadrilateral because $\angle{ATC} =\angle{AMB} =\angle{ADC} $) So $A$ is the center of spiral similarity that takes $BM$ to $CT$, so it will takes $BC$ to $MT$ $\Longrightarrow \triangle ABC \sim \triangle AMT$ $$ \Longrightarrow \angle{AMT}=\angle{ABC} =\angle{DMC} ...(1)$$Since the perpendicular bisector of $AC$ bisects $\angle{AMC}$ so from $(1)$ it bisects $\angle{DMT}$ and since the reflection of $D$ in the perpendicular bisector of $AC$ lies on $(DACT)$ so it is obvious that $T$ is the reflection of $D$ in the perpendicular bisector of $AC$. $$\Longrightarrow \triangle CAD \simeq \triangle ACT$$But since $\triangle ACT \sim \triangle ABM$ so we have $\triangle CAD \sim \triangle ABM$ $$\Longrightarrow \frac{AD}{DC} =\frac{MB}{MA}$$And then we also have $\triangle ABC \sim \triangle CMD \Longrightarrow \frac{AB}{BC} =\frac{MC}{MD}$ As desired... $\blacksquare$

Let $DM$ intersect $BC$ at $P$ and $BM$ intersect $AD$ at $Q$. Let $B_0$ be on the same side as $B$ of $AC$ such that $\triangle AB_0C\sim \triangle AMD$. Note that $\angle MPC=\angle DMC-\angle MCP=\angle MAP$ so $MABP$ is cyclic. Additionally, $\angle DQM=\angle AMB-\angle MAD=\angle MCD$ so $MDQC$ is cyclic. $~$ Now, $\angle AB_0C=\angle AMD=180^\circ-\angle AMP=\angle ABP=\angle ABC$, so $ABB_0C$ is cyclic. Thus, $A$ is the Miquel Point of quadrilateral $MDCB_0$. We have \[\angle AMB_0=\angle ADC=180^\circ-\angle QDC=180^\circ-\angle QMC=\angle BMC\]By symmetry across the perpendicular bisector of $AC$, $ABB_0C$ is an isosceles trapezoid, so $\triangle ABC\sim \triangle DMA$. We also have $\angle ADC=\angle BMC$ and $\angle DCA=\angle MB_0A=\angle MBC$ so $\triangle MBC\sim \triangle DCA$. In particular, \begin{eqnarray*} \frac{DM}{AM}=\frac{AB}{BC}&\implies& AB \cdot CM = BC \cdot MD \\ \frac{BM}{MC}=\frac{CD}{AD}&\implies& BM \cdot AD = MA \cdot CD \end{eqnarray*}as desired.