Hint: If your four points are $a, b, c, d$, then the midpoints, in order around the quad, are
$$
p = \frac{1}{2}(a+b), q = \frac{1}{2}(b+c), r = \frac{1}{2}(c+d), s = \frac{1}{2}(d+a).
$$
For $pqrs$ to be a parallelogram, you need the edge from $p$ to $q$ to have the same direction vector as the edge from $s$ to $r$; you need a similar thing to hold for the edges from $q$ to $r$ and $p$ to $s$.
What's the direction vector of the edge from $p$ to $q$? Can you express it in terms of $a, b, c, d$?
Draw a parallegram, label the corners from lower left, counter clockwise:
A(lower left), B(lower right),C(Upper right),D(Upper left).
Let $AB$ be vector $\vec b$, $AD$ be vector $\vec a$.
Diagonal $AC$ : $\vec a + \vec b$.
Diagonal $BD$ : $\vec a - \vec b.$
Label the point of intersection of the diagonals $O$.
Consider $\triangle$ $ABO$:
$\alpha (\vec a + \vec b) = \vec b + \beta (\vec a - \vec b), \alpha, \beta \in \mathbb{R}$.
Collect coefficients of $\vec a$ and $\vec b$:
$(\alpha - \beta) \vec a + (\alpha + \beta - 1) b = 0$
Since $\vec a$ and $\vec b$ are independent (not collinear) it follows that the coefficients of $\vec a$ and $\vec b$ must vanish.
1) $\alpha - \beta = 0$, and
2) $\alpha + \beta - 1 = 0.$
Adding and subtracting:
$\alpha = 1/2$, $\beta = 1/2.$
Best Answer
Given conditions are
$\vec{CM} = \vec{MA}$
$\vec{MB} = \vec{OM}$
It can be seen using the triangle law of addition that
$\vec{CB} = \vec{CM} + \vec{MB}$
$\vec{OA} = \vec{OM} + \vec{MA}$
It can clearly be seen that $\vec{CB} = \vec{OA}$ (substitute the first 2 equations into any one of the last 2 equations). And by the theorem "If two sides of a quadrilateral are parallel and equal, the quadrilateral is a parallelogram (see the $4^{th}$ characterisation here)", the result is proved.