If the medians of a `triangle`ABC intersect at G, show that
`ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)`
`=(1)/(3)ar(triangleABC)`.

Explanation

Given In `DeltaABC, AD, BE " and " CF` are medians and intersect at G.

To prove `ar (DeltaAGB) = ar (DeltaAGC) = ar (DeltaBGC) = (1)/(3)ar (DeltaABC)`
Proof We know that, a median of a triangle divides it into two triangles of equal area.
In `DeltaABC, AD` is a median.
`therefore" "` `ar (DeltaABD) = ar (DeltaACD)" "` ...(i)
In `DeltaBGC, GD` is a median.
`therefore" "` `ar (DeltaGBD) = ar (DeltaGCD)" "` ...(ii)
On substracting Eq. (ii) from Eq. (i), we get
`ar (DeltaABD) - ar (DeltaGBD) = ar (DeltaACD) - ar (DeltaGCD)`
`rArr" "` `ar (DeltaAGB) = ar (DeltaAGC)" "` ...(iii)
Similarly,
`ar (DeltaAGB) = ar (DeltaBGC)" "` ...(iv)
From Eqs. (iii) and (iv),
`ar (DeltaAGB) = ar (DeltaBGC) = ar (DeltaAGC)" "` ...(v)
Now,
`ar (DeltaABC) = ar (DeltaAGB) + ar (DeltaBGC) + ar (Delta AGC)`
`rArr" "` `ar (DeltaABC) = ar (DeltaAGB) + ar (DeltaAGB) + ar (DeltaAGB)" "` [from Eq. (v)]
`rArr" "` `ar (DeltaABC) = 3 ar (Delta AGB)`
`rArr" "` `ar (DeltaAGB) = (1)/(3)ar (DeltaABC)" "` ...(vi)
From Eqs. (v) and (vi),
`ar (DeltaBGC) = (1)/(3) ar (DeltaABC)`
and `" " ar (Delta AGC) = (1)/(3) ar (DeltaABC)" "` Hence proved.