an easy nonrigorous way is just to note that the problem statement implies that the expression is constant for any choice of O, then choose something convenient.

You can make it rigorous by showing that it's invariant...

OK // AC, and MN // BC $\Rightarrow$ ONKC is a paralellogram. Thus, $CN$ = $OK$ $\angle EOK = \angle BAC, \angle OKE = \angle ACB \Rightarrow \Delta ABC$ is similar to $\Delta OEK$ \[ \Rightarrow \dfrac{CN}{CA}=\dfrac{OK}{CA}=\dfrac{EK}{BC} \] Because AF // OD, and OF // AD, AFOD is a parallellogram. Therefor, AF = OD. Because $\angle ODN = \angle BAC$, and $\angle DON = \angle ABC, \Delta ODN$ is similar to $\Delta ABC$ \[ \Rightarrow \dfrac{AF}{AB}=\dfrac{OD}{AB}=\dfrac{ON}{BC}=\dfrac{KC}{BC} \] \[ \dfrac{AF}{AB}+\dfrac{BE}{BC}+\dfrac{CN}{CA}=\dfrac{BE+EK+KC}{BC}=1 \]

Through an internal point $O$ of $\Delta ABC$ one draws $3$ lines, parallel to each of the sides, intersecting in the points shown on the picture. Find the value of $\frac{|AF|}{|AB|}+\frac{|BE|}{|BC|}+\frac{|CN|}{|CA|}$.