What's the measure of the segment $PD$ in the rhombus below? For reference : Let $M$ and $N$ be the midpoints of the sides $AB$ and $AD$ of a rhombus $ABCD$.
Calculate $PD$, if $MP= 6, (P = MD \cap CN$)
My progress:

G is centroide
$\triangle ABD\\
GM = \frac{2M}{3}\\
GM = 2DG = 2(GP+DP)\\
MD = 3DG\\
GM = 6 -GP\\
\therefore 6 - GP = 2GP+2DP \implies \\
2DP = 6 - 3GP$
There is one more relationship to finish...
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As $\triangle NGM \sim \triangle BGD$, $DG = 2 GM$ as you mentioned. Also, $GH = \dfrac{AH}{3}$.
In $\triangle AGD$, traversal $NC$ intersects $AD$ and $GD$ internally and $AG$ externally. Applying Menelaus's theorem,
$ \displaystyle \frac{AC}{CG} \cdot \frac{GP}{PD} \cdot \frac{DN}{NA} = 1$
$ \displaystyle \frac{2 AH}{4AH/3} \cdot \frac{GP}{PD} = 1 \implies PD = \frac{3}{2} GP$
and we obtain $ ~ 5 GP = 4GM$
As $MP = 6, GP = \dfrac{8}{3} \implies PD = 4$
