Seems easy

[SANZEED]

Let $ABC$ be triangle. Cevian $AA_1$ intersect $BC$ at $A_1$ such that $BA_1:A_1C=2:1$. If $CC_1$ is a median,then in what ratio will it intersect $AA_1$?

[SANZEED]

I have a nice and very simple proof by "Mass Point Geometry" ( See http://en.wikipedia.org/wiki/Mass_point_geometry ). Is it a Euclidean technique? i am not sure whether it is Euclidean or not.

[*Mahi*]

As far as I know, "Mass Point geometry" or something similar does not occur in "Elements" by Euclid, and it isn't any direct consequences of anything stated in that. So, I do not think that this can be called Euclidean.

[sakibtanvir]

Sorry,but pls clarify what is meant by "ratio of intersection".If the intersection point is $R$,then what the question asks, $\frac{CR}{C_{1}R}$ or $\frac{AR}{A_{1}R}$ ?

[photon]

let $O$ is the intersecting point ,
${\frac {OA}{OA_1}}={\frac {\Delta AOB}{\Delta A_1OB}}={\frac {2\Delta AOC_1}{2\Delta A_1OC}}={\frac {\frac {1}{2}. OA.OC_1 sin \angle AOC_1}{\frac {1}{2}.OA_1.OCsin \angle A_1OC}}={\frac {OA.OC_1}{OA_1.OC}}$
$\Rightarrow OC_1=OC$ ; $\therefore \Delta BOC_1=\Delta BOC \Rightarrow \Delta AOC_1=3\Delta A_1OC $
$OA:OA_1= \Delta AOC_1:\Delta A_1OC=3:1$