If the medians of a ` A B C` intersect at `G ,` show that `a r( A G B)=a r( A G C=a r( B G C)=1/3a r( A B C)dot` GIVEN : ` A B C` such that its medians `A D ,B E` and `C F` intersect at `Gdot` TO PROVE : `a r( A G B)=a r( B G C)=a r(C G A)=1/3A R( A B C)`

Explanation

Given: A `∆ABC` and its medians `AD`, `BE` and `CF` intersect at `G`.
To prove : `ar(/_\AGB)=ar(/_\AGC)=ar(/_\BGC)=1/3ar(/_\ABC)`.
Proof: A median of a triangle divides it into two triangles of equal area.
In `/_\ABC`, `AD` is the median.
`:.ar(/_\ABD)=ar(/_\ACD)` ...(i)
In `/_\GBC`, `GD` is the median.
`:.ar(/_\GBD)=ar(/_\GCD)` ...(ii)
From (i) and (ii), we get:
`ar(/_\ABD)-ar(/_\GBD)=ar(/_\ACD)-ar(/_\GCD)`
`:.ar(/_\AGB)=ar(/_\AGC)`.
Similarly, `ar(/_\AGB)=ar(/_\AGC) =ar(∆BGC)` ....(iii)
But, `ar(/_\ABC)=ar(/_\AGB)+ar(/_\AGC)+ar(/_\BGC)=3ar(/_\AGB)` [Using (iii)]
`:.ar(/_\AGB)=1/3ar(/_\ABC)`
Hence, `ar(/_\AGB)=ar(/_\AGC)=ar(/_\BGC)=1/3ar(/_\ABC)`.