Geometry problems that me and my squad cannot solve. This quarantine, we decided to make some geometrical research, we got stucked on this:
In a $\triangle ABC$ triangle, let the Cevians $AL$, $BM$, $CN$ concur on a point $O$.
Show that: $\frac{OL}{AL}+\frac{OM}{BM}+\frac{ON}{CN}=1$
I am using the Ceva equality to get that 1, but we are not sure about that. Also, what does this theorem mean in terms of dividing the cevians in parts ?
Any hint/solution is welcome. Thank you !
Hint: The idea is to use ratio of areas of triangle. Thus: $\dfrac{OL}{AL} = \dfrac{S_{OBC}}{S_{ABC}}$. Repeat this with the other two fractions and add up...