How to prove that $ABCD$ is a parallelogram? 
Let $ABCD$ be a quadrilateral. Let $E$ and $F$ be midpoint of $AB$ and $BC$. The lines $DE$ and $DF$ intersect $AC$ at $M$ and $N$ respectively.  Suppose that $AM$ $=$ $MN$ $=$ $MC$. Prove that $ABCD$ is a parallelogram.

What I know is $EM$ are parallel to $BN$ and also $EM$ length is half of $BN$.  In the same way with $NF$ and $BM$. I also tried Appolonius Theorem but failed. Hint please. Thank you very much!
 A: It suffices show that $AC$ bisects $BD$.  Let $X$ be the intersection between $AC$ and $BD$.
Since $E$ and $F$ are midpoints of $AB$ and $BC$, $BX$ is bisected by $EF$.  If $Y$ is the point of intersection between $BD$ and $EF$, then $|XY|=|BY|$, so that $|XY|=\dfrac{|BX|}{2}$.
Let $\ell:=|AC|$.  Then, $|MN|=\dfrac{\ell}{3}$.  Because $|EF|=\dfrac{|AC|}{2}=\dfrac{\ell}{2}$, we conclude that $\dfrac{|MN|}{|EF|}=\dfrac{2}{3}$.  Since $MN\parallel EF$, we conclude that the line $\dfrac{|DX|}{|DY|}=\dfrac{2}{3}$, implying $$|XY|=|DX|-|DY|=\dfrac{|DX|}{2}\,.$$  That is, $|DX|=2\,|XY|=|BX|$.  Thus, $AC$ bisects $BD$, as desired.

  To add the final piece of argument, note that $DE$ and $AX$ are medians of the triangle $ABD$.  Therefore, $M$ is the centroid of the triangle $ABD$, whence $\dfrac{|AM|}{|MX|}=2$.  This shows that $X$ is the midpoint of $AC$.  Hence, $BD$ also bisects $AC$.

A: Since $NF$ is midlle line in $BCM$ we have $MB||NF$ and so $DN||MB$
Similary $NB||DM$ and so $BMDN$ is parallelogram which means $MN$ and $BD$ halves each other. But then $AC$ and $BD$ halves each other and so $ABCD$ is parallogram.
