The Theorems of Ceva and Menelaus

Question

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 $DO:OM$?

Answer 1

Let [.] denote areas. Since the triangles ADN and AMN share the same base AN,

$$\frac{DO}{OM} = \frac{[ADN]}{[AMN]} = \frac{\frac12[ABCD]}{\frac29[ABN]} = = \frac{\frac12[ABCD]}{\frac29\cdot\frac4{2\cdot9}[ABCD]} = \frac{81}{8}$$

Similarly,

$$\frac{AO}{ON} = \frac{[DAM]}{[DNM]} = \frac{[DAM]}{[ABCD]-[DAM]-[NBM]-[DNC] } $$ $$= = \frac{\frac2{2\cdot9}[ABCD]} {(1-\frac2{2\cdot9}- \frac7{2\cdot9}\cdot\frac49- \frac5{2\cdot9})[ABCD]} = \frac{18}{71}$$

Answer 2

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: $${AO\over ON} = {AD\over QN} = {9\over t+4} ={18\over 71}$$

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Let $CP = x$. Again by Thales: $${x\over x+9}={CP\over DP} = {CN\over AD} = {5\over 9}\implies x=45/4$$

so

$${DO\over OM} = {DP\over AM} = {9+x\over 2}={81\over 8 } $$