Question

If the medians of a ΔABC intersect at G show that ar (AGB) = ar (AGC) =ar (BGC) = 1/3 ar (ABC)

Answer 1

2)in the fig.consider tri.ABC
AD is the median,so ar.ABD=arACG                                        (1)
now,consider tri.BGC
GD is the median so, ar.BGD=CGD                                          (2)
subtracting 2 from 1,we get,
ar.AGB=AGC
llly,Ar. BGC=AGC
i.e.Ar[AGB=AGC=BGC]
and ar.[AGB+G=BGC+AGC=ABC]
hence,the areas of all the triangles so formed is 1/3 Ar of ABC