Area of a parallelogram using similar triangles 
$ABCD$ is a parallelogram and point $M$ lies on $AB$ such that $AM:MB=2:3$. If $DM \cap AC=N$ and the area of $\triangle ADN=a$, I should find the area of the parallelogram $ABCD$.


Let $AD=BC=b$ and $NN_1\perp AD=h_1, BB_1 \perp AD = h_2$. We have $S_{\triangle ADN}=\dfrac{AD.NN_1}{2}=\dfrac{b.h_1}{2}$ and $S_{ABCD}=AD.BB_1=b.h_2$. This is my idea but I can't approach the problem further. Maybe we can try to find $\dfrac{h_1}{h_2}$. Can you give a hint on how to continue? Thank you in advance!

Hint: $\Delta ANM$ is similar to $\Delta DNC$, so we need to find the area of $\Delta DNC$ in terms of $a$.
Let $Area(ABCD) = A$
$$Area(AND) + Area(CND) = \dfrac{A}{2}$$
$$\frac{Area(ANM)}{Area(CND)} = \bigg(\frac{AM}{CD}\bigg)^2 = \bigg(\frac{2}{5}\bigg)^2 = \frac{4}{25}$$
Also, 
$$Area(AND) + Area(ANM) = \dfrac{A}{5}$$
So, assuming $Area(CND) = x$, 
$$\therefore a + x = \frac{A}{2} \text{ and } a + \frac{4}{25}x = \frac{A}{5}$$
Now you can easily find both $a$ and $x$ in terms of $A$.
($a = \frac{A}{7}$ and $x = \frac{5}{14}A$)
Also, to answer your original question, yes you can find $\dfrac{h_1}{h_2}$, but in the end, it boils down to finding some similarity relations.
