Given: In ∆ABC, medians AD, BE and CF intersect at G.
To prove:
ar (∆AGC) = ar (∆AGB)
= ar (∆BGC) = \(\frac { 1 }{ 3 }\) ar (∆ABC)
Proof: We know that median of a triangle divides it into two triangles of equal areas.
In ∆ABC, AD is a median
ar (∆ABD) = ar (∆ACD) …(i)
In ∆CBG, GD is a median
ar (∆GBD) = ar (∆GCD) …(ii)
On subtracting equation (ii) from equation (i), we get
ar (∆ABD) – ar (∆GBD) = ar (∆ACD) – ar (∆GCD)
⇒ ar (∆AGB) = ar (∆AGC)
Similarly,
ar (∆AGB) = ar (∆BGC)
ar (∆AGB) = ar (∆AGC) = ar (∆BGC) …(iii)
But ar (∆ABC) = ar (∆AGB) + ar (∆AGC) + ar (∆BGC) = 3 ar (∆AGB) [from equation (iii)]
ar (∆AGB) = \(\frac { 1 }{ 3 }\) ar (∆ABC)
Hence, ar (∆AGB) = ar (∆AGC) = ar (∆BGC)
= \(\frac { 1 }{ 3 }\) ar (∆ABC)
Hence proved.
