In the figure below, given a triangle
ABC of area S, the medians BM and CN meet at G. Let be BD = 2 CD, the cevian AD meets BM and CN at E and F respectively.
Area of triangle EFG is S_{1}. Prove that:
GE/EM = 2/3, GF/CF = 1/3 and
S_{1} = S/60.
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