In a given triangle $ABC$ lets choose inside side $AC$ points $P$, $Q$ and inside side $BC$ points $R$,$S$ so, that $|AP|=|PQ|=|QC|$, $|BR|=|RS|=|SC|$
Next denote point $E$ as intersection of diagonals in trapezoid $ABRP$, point $F$ as intersection of diagonals in trapezoid $PRSQ$ and point $G$ as intersection of diagonals in trapezoid $ABSQ$. Prove for any triangle, that points $E$, $F$ and $G$ lies on median of triangle $ABC$ from vertex $C$.
Triangle sketch:
It is enough to show that $F$ lies on median from $C$. For the others points the solution is exactly the same.
The triangles $CQS$ and $CPR$ and $CAB$ are all similars (by SAS case) then the median from $C$ is also median of the triangles $CQS$ and $CPR$ and the it meets $QS$ and $PR$ in the midpoint.
Let's call them $M_1,M_2$. In order to prove that $F$ is on the median it is enough to prove that $\overline{FM_1}$ and $\overline{FM_2}$ lies on the same line.
$QS \parallel PR$ then $\angle M_1QF =\angle FRM_2$ and also $\angle QM_1F=\angle FM_2P$.
That give us $\angle M_1FQ=\angle RFM_2$ and then $\overline{FM_1}$ and $\overline{FM_2}$ lies on the same line.
The concepts your problem uses are invariant under affine transformations. So without loss of generality you can restrict yourself to the simple case
$$C=(0,0),\; Q=(1,0),\; P=(2,0),\; A=(3,0),\; S=(0,1),\; R=(0,2),\; B=(0,3)$$
There you can explicitely give the coordinates of the points of intersection, and the equation of the median, and see that the points lie on the median.