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  1. $ACB=ACD$
  2. $BAC=ADC$
  3. $MB=MA$

$Prove \angle MCB=\angle MBD$

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Let $E\ (\not=C)$ be a point on the line $CM$ such that $MC=ME$, and let $F$ be the intersection point of $BD$ with $CE$. Since the quadrilateral $AEBC$ is a parallelogram, we have $\angle{EBA}=\angle{BAC}$ and $\angle{AEB}=\angle{ACB}$.

We know that $\triangle{ABE}$ and $\triangle{BAC}$ are congruent, and that $\triangle{BAC}$ and $\triangle{ADC}$ are similar.

So, if we set $BA=a,AE=b,EB=c$, then $$BC=b,\quad AC=c$$ and $$AD:BA=AC:BC\implies AD:a=c:b\implies AD=\frac{ac}{b}$$

So, we have $$\angle{EBC}=\angle{DAB}\quad \text{and}\quad BE:BC=AD:AB$$ from which we have that $\triangle{EBC}$ and $\triangle{DAB}$ are similar.

Therefore, $\angle{MBD}=\angle{ABD}=\angle{BCE}=\angle{MCB}$.

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Just for fun, a different solution. Let $N$ be the midpoint of $AD$ and let $L$ be the intersection point between $CN$ and $BD$. Then triangles $CND$ and $CMA$ are similar, so $\angle CND = \angle CMA$. Therefore $\angle CNA + \angle CMA = \pi - \angle CND + \angle CMA = \pi$. Therefore the four points $C, M, A, N$ lie on a common circle. Therefore $\angle CNM = \angle CAM = \angle CAB$. But since $M$ and $N$ are midpoints of $AB$ and $AD$ respectively, the segment $MN$ is parallel to $BD$ (and half of it in length). Therefore $\angle CLB = \angle CNM = \angle CAB$ which means that the four points $C, L, A, B$ lie on a common circle and therefore $\angle NCA = \angle LCA = \angle LBA = \angle ABD$. But since triangles $NCA$ and $MCB$ are similar, we conclude that $\angle MBD = \angle ABD = \angle NCA = \angle MCB$.

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