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

2
by shankaraJani

2015-03-26T17:39:00+05:30
Construction : Draw a perpendicular from A and G and name it as M and N respectively.
We know that medians of a triangle intersect at a ratio of 2:1.
Thus, AG/GD = 2/1
or, AG + GD/GD = 2+1/1
or, AD/GD = 3/1 .......................... (i)

Now, in triangles AMD and GND,
<AMN = <GND
Thus triangle AMD ~ GND (AA)

So, we get AM/GN = AD/GD
or,  AM/GN = 3/1 [From (i)]
Next ar(ABC) : ar(BGC)
= 1/2 x BC x AM : 1/2 x BC x GN
= AM : GN
= 3 : 1
or, ar(GBC) = 1/3ar(ABC)

Hope that helps !!
2015-03-26T17:44:23+05:30
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