Two parallelograms and a centroid 
A parallelogram $ABCD$ is given. The points $M_1$ and $M_2$ are on
   $AC$ and $M_3$ and $M_4$ are on $BD$ such that $AM_1 = CM_2 =
 \frac{1}{3}AC$ and $BM_3=\frac{1}{3}BD$. If $M_1M_3M_2M_4$ is a
   parallelogram, show that $M_4$ is the centroid of $\triangle ACD$.

I started with $AM_1=CM_2=M_1M_2=\frac {1}{3}AC=2x$. If $M_1M_2 \cap M_3M_4=O$, $M_1O=OM_2=x$. I don't know how to continue. I have to show that $\frac {DM_4}{M_4O}=\frac{2}{1}.$ Thank you in advance! 
 A: Due to similarity of triangles, it's easy to show $M_1M_3$ is parallel to $AB$. If $M_1M_3M_2M_4$ is parallelogram, $M_2M_4$ is also parallel to $AB$. Then you show $M_2M_4=M_4D=\frac 13 BD$. Also, in the parallelogram the diagonals intersect in such a way that $OM_4=\frac 12 M_2M_4$, and that is the thing you wanted to show.
Here is an alternative: Similarly to what you have done before, $BM_3=2y$, then $OM_3=OM_4$ and $BO=\frac 12 BD$. Therefore $\frac 12 BD=BO=BM_3+M_3O=\frac 13 BD+M_3O$ yields $M_3O=\frac 16 BD$. Now due to congruent triangles $\triangle BAC\equiv\triangle DCA$, you get $DO=\frac 12 BD$, and since $OM_4=OM_3=\frac 16 BD$, you get $DM_4=DO-OM_4=\frac 12 BD-\frac 16 BD=\frac 13 BD=2\frac 16 BD=2OM_4$ 
A: Do you know that the centroids divides each median in the ratio $2:1$?  If so, it's easy to show that $M_3$ is the centroid of $ABC$.  Now use congruence.
A: 
$M_1M_2M_3M_4$ is a parallogram
$\triangle M_1OM_3 \cong \triangle M_2OM_4\\
OM_3 = OM_4$
$ABCD$ is a parellogram
$BO= OD\\
DM_4 = BM_3$
If $M_3$ cuts $OB$ at a ratio 1:2 so $M_4$
$DO$ is a median of $AC$
The centroid of a triangle is found on the median 2/3 of the way from the vertex.
A: If we use (position) vectors then we have $$\vec{M_1} = {2\over 3}\vec{A}+{1\over 3}\vec{C}$$
$$\vec{M_2} = {2\over 3}\vec{C}+{1\over 3}\vec{A}$$ 
$$\vec{M_3} = {2\over 3}\vec{B}+{1\over 3}\vec{D}$$ 
so $$\vec{M_4} = \vec{M}_1+\vec{M}_2-\vec{M}_3 $$
$$= \vec{A}+\vec{C}-{2\over 3}\vec{B}-{1\over 3}\vec{D}$$
$$= \vec{A}+\vec{C}-{2\over 3}(\vec{A}-\vec{D}+\vec{C})-{1\over 3}\vec{D}$$
$$= {1\over 3}(\vec{A}+\vec{D}+\vec{C})$$ 
and we are done.

Say we have some point $O$, name it origin. Then for each point $X$ we can make a (position) vector with respect to $O$ which start at $O$ and ends at $X$ and we mark it simply by $\vec{X}$. Where is $O$ is up to you and many times it is quite uninportant where it is (like in this case). However, we have for each vector $\vec{XY}=\vec{Y}-\vec{X}$
Point $T$ is a centroid of a triangle $ABC$ iff $\vec{T}= {1\over 3}(\vec{A}+\vec{B}+\vec{C})$ 
$ABCD$ is a paralelogram iff $\vec{AB}=\vec{DC}$, i.e. $\vec{B}- \vec{A}=\vec{C}-\vec{D}$ 
