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

2
by shankaraJani

Log in to add a comment

by shankaraJani

Log in to add a comment

AD is the median,so ar.ABD=arACG

now,consider tri.BGC

GD is the median so, ar.BGD=CGD

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