<s>[INST] The Theorems of Ceva and Menelaus 
In the parallelogram $ABCD$ $M\in AB, AM:MB=2:7$ and $N\in BC, BN:NC=4:5$. If $DM\cap AN=O$, calculate $AO:ON;DO:OM$.


For the first ratio let $BC \cap DM=Q$. By Menelaus' theorem $\dfrac{AO}{ON}\cdot\dfrac{NQ}{QB}\cdot\dfrac{BM}{MA}=1 \Rightarrow \dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}$. If we look at the similar triangles $\triangle BMQ \sim AMD \Rightarrow \dfrac{BQ}{AD}=\dfrac{BQ}{BC}=\dfrac{BM}{AM}=\dfrac{7}{2}$ or $BQ=\dfrac{7}{2}BC$. Now, $\dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}=\dfrac{\dfrac{7}{2}BC}{BQ+BN}\cdot\dfrac{2}{7}=\dfrac{\dfrac{7}{2}BC}{\dfrac{7}{2}BC+\dfrac{4}{9}BC}\cdot\dfrac{2}{7}=\dfrac{63}{71}\cdot\dfrac{2}{7}=\dfrac{18}{71}$. Thales theorem tells us that
$\dfrac{AO}{ON}=\dfrac{DO}{OQ}=\dfrac{18}{71}$. 
How to find $DO:OM$?
 [/INST] Let $QB= t$. By Thales: $${DC\over MB}  = {QC\over QB} \implies {9\over 7}  = {t+9\over t}\implies t=63/2$$
so again by Thales:
$${AO\over ON} = {AD\over QN} = {9\over t+4} ={18\over 71}$$

Let $CP = x$. Again by Thales: $${x\over x+9}={CP\over DP} = {CN\over AD} = {5\over 9}\implies x=45/4$$
so
$${DO\over OM} =  {DP\over AM} = {9+x\over 2}={81\over 8 } $$
</s>