Centroid of Triangle | SMO, 2009 | Problem 1

If we got stuck in this problem then we can start this problem by applying Menelaus's Theorem.

It states : if a line intersects \(\triangle ABC\) or extended sides at points D, E, and F, the following statement holds: \(\frac {AD}{BD} \times \frac {BE}{EC} \times \frac {CF}{AF} = 1\)

Again let D is the mid point of AC. As \(\frac {BM}{MA} + \frac {CN}{NA} = 1\) then \(\frac {CN}{NA}<1\) where N lies in the line segment CD.From the picture above we can see g is the intersection point between two lines BD and MN. So if we apply Menelaus's Theorem we get :

\(\frac {DG}{GB} . \frac {BM}{MA} .\frac {AN}{ND} = 1\)

Now try the rest of the problem.....................................

After the 1st hint again, \(\frac {BG}{GD} = \frac{BM}{MA} . \frac {AN}{ND} = ( 1 - \frac {CN}{NA}). \frac {AN}{ND}\)

= \(\frac {NA - CN}{ND} = \frac {(2CD - CN) - CN}{ND}\)

=\(\frac {2 ND}{ND}\) = 2

Thus G is the centroid .(Proved)