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...
$

Say $CN$ intersects $BD$ at $R$.
$BH$ and $CM$ are medians of $\triangle ABC$.
Therefore $Q$ is the centroid of $\triangle ABC$. So $BQ:QH=2:1$. From there we can see $BQ=QR=RD$.
Also from midpoint theorem to $\triangle ABD$ we found $MN=\dfrac12BD$.
Hence $MN=\dfrac32RD$.
Now, $\triangle MNP\sim\triangle DRP\implies\dfrac{MN}{RD}=\dfrac{MP}{PD}=\dfrac32\implies PD=4.$
