What's the measure of the segment $PD$ in the rhombus below?

Question

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... $

Answer 1

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, $\triangle MNP\sim\triangle DRP\implies\dfrac{MN}{RD}=\dfrac{MP}{PD}=\dfrac32\implies PD=4.$

Answer 2

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 $MP = 6, GP = \dfrac{8}{3} \implies PD = 4$

Answer 3

One can use a simple approach using equations of lines. Let $H$ be the origin, $C$ be at $(2\alpha,0)$ and $B$ at $(0,2\beta)$. Then $M=(-\alpha,\beta)$, $N=(-\alpha,-\beta)$, and $D=(0,-2\beta)$. Then we write the equation for $MD$: $$\frac{x-x_M}{x_D-x_M}=\frac{y-y_M}{y_D-y_M}\\\frac{x+\alpha}{\alpha}=\frac{y-\beta}{-2\beta-\beta}\\y=-\frac{3\beta}\alpha x-2\beta$$
Similarly, for $NC$: $$\frac{x-x_N}{x_c-x_N}=\frac{y-y_N}{y_C-y_N}\\\frac{x+\alpha}{3\alpha}=\frac{y+\beta}{\beta}\\y=\frac{\beta}{3\alpha} x-\frac23\beta$$

When we write the intersection (point $P$), we get: $$-\frac{3\beta}\alpha x_P-2\beta=\frac{\beta}{3\alpha} x_P-\frac23\beta\\ x_P=-\frac{4}{10}\alpha$$ Now draw perpendiculars from $P$ and $M$ to $BD$. You get similar triangles. You can then write $$\frac{MD}{PD}=\frac{6+PD}{PD}=\frac{\alpha}{\frac4{10}\alpha}=\frac{10}4$$ This yields $PD=4$