$ABCD$ 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!
Best Answer
$$\frac{S_{\Delta AMN}}{a}=\frac{NM}{DN}=\frac{AM}{DC}=\frac{2}{5}.$$ Thus, $$S_{\Delta ADM}=S_{\Delta ADN}+S_{\Delta AMN}=a+\frac{2}{5}a=\frac{7}{5}a.$$ Now, $$\frac{S_{\Delta ADB}}{S_{\Delta ADM}}=\frac{AB}{AM}=\frac{5}{2},$$ which gives $$S_{ABCD}=2S_{\Delta ADB}=2\cdot\frac{5}{2}\cdot\frac{7}{5}a=7a.$$