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I was looking at this question and managed to find expressions for most of the sides (I also found the vector OM, but I am unsure what to do next). I am thinking that you need to set up some simultaneous equation with ON or AP, but can anyone tell me how they would solve this?

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  • $\begingroup$ you could just use menelaus's theorem for OMB. $\endgroup$ Commented May 6, 2019 at 21:37
  • $\begingroup$ How would that work? $\endgroup$ Commented May 7, 2019 at 5:48

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Let $\overrightarrow{ON}=k\mathbf{b}$, we have: $$ \overrightarrow{AP}={4\over7}{\mathbf{a}+\mathbf{b}\over2}- \mathbf{a}={2\over7}\mathbf{b}-{5\over7}\mathbf{a}, \quad \overrightarrow{AN}=\overrightarrow{ON}-\mathbf{a}=k\mathbf{b}-\mathbf{a}. $$ But $\overrightarrow{AN}=t\overrightarrow{AP}$ for some $t$, that is: $$ k\mathbf{b}-\mathbf{a}={2\over7}t\mathbf{b}-{5\over7}t\mathbf{a}. $$ We thus get: $t={7\over5}$ and $k={2\over5}$. Finally: $$ ON:NB=k:(1-k)=2:3. $$

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