Similarity Ratios in triangle

Question

Triangle $ABC$ is given. Let $M$ and $N$ be points on its sides $BC$ and $AC$ respectively, such that $BM/MC$ = $1/3$ and $AN/NC$ = $1/5$. If $O$ is the intersection point of $AM$ and $BN$, find the ratio $AO/OM$.

I drew BN and AM and I see that there is a similar triangle relation there; $\triangle ANO \sim \triangle BMO$. So I got the ratio:

$\frac{AO}{OM} = \frac{AN}{BN}= \frac{3}{5} NC/MC$.

If you take a triangle now, with sides: $BM=1, MC = 3, AN = 1, NC = 5$, it messes everything up.

How do I approach this?

Answer 1

By Menelaus' theorem, we have $$\frac{CB}{BM}\cdot\frac{MO}{OA}\cdot\frac{AN}{NC}=1.$$ So, $$\frac{AO}{OM}=\color{red}{\frac 45}.$$

Answer 2

I got an answer too. Inspired from Jack D'Aurizio's answer in my other question.

Van Obel and Ceva's theorem is used.

The problem gives: $BM/MC = 1/3, AN/NC = 1/5$

Let there be a point $F$ on $AB$ such that $CF$ intersects at $O$ as well. Then, by Ceva:

$\frac{AF}{FB} \cdot \frac{BM}{MC} \cdot \frac{CN}{NA} = 1 \implies \frac{AF}{FB} = \frac{3}{5}.$

By Van Obel's theorem, $\frac{AO}{OM} = \frac{AF}{FB} + \frac{AN}{NC} \implies \frac{AO}{OM} = 1/5 + 3/5 = \frac{4}{5}$ as @mathlove got.