Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Centroid of Triangle.
Let M and N be points on sides AB and AC of triangle ABC respectively. If \(\frac {BM}{MA} + \frac {CN}{NA} = 1\) . Can we show that MN passes through the centroid of ABC?
2D - Geometry
Triangle
Menelaus's Theorem
Answer: Yes , we can.
Singapore Mathematics Olympiad
Challenges and Thrills - Pre College Mathematics
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)
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