Parallelogram and vectors

In the parallelogram $ABCD$ points $M$ and $N$ are the midpoints of $BC$ and $CD$, respectively. Point $P$ is such that $AMPN$ is a parallelogram. Show that $C\in AP$. I am supposed to use vectors. The thing that characterize the point $P$ is that the quadrilateral $AMPN$ is a parallelogram. This means $\vec{AM}=\vec{NP}$ and $\vec{AN}=\vec{MP}$. I am not sure how we can further use that. Thank you in advance!

Solution 1:

Set $A$ as the origin (this is just to make calculations easier, I can post the solution without this assumption if required).

$ABCD$ is a parallelogram, hence $C=B+D$.
$M$ is the midpoint of $BC$, hence $M=\dfrac{B+C}{2}=\dfrac{2B+D}{2}$.
$N$ is the midpoint of $CD$, hence $N=\dfrac{C+D}{2}=\dfrac{B+2D}{2}$.
$AMPN$ is a parallelogram, hence $P=M+N=\dfrac{2B+D}{2}+\dfrac{B+2D}{2}=\dfrac{3}{2}(B+D)=\dfrac{3}{2}C$.
Therefore, $C\in AP$.