Question

In triangle ABC the medians AD,BE,CF intersect at G. Prove that ar(AGB) = $\frac{1}{3}$ ar(ABC)

Answer 1

A median bisects the triangle into two equal parts. Based on this, we can write,
Area of triangle ABC = 2 ar ADB = 2 ar BEC = 2 ar ADC
(all the following are areas)
$\to$ ADB = BEC = ADC
$\to$ AGB + GBD = AGC + GCD ; But, GBD = GCD (because D is midpoint of BC in triangle GCD)
$\to$ AGB = AGC
Similarly it can be proved that AGB = BGC.
So, we have AGB = AGC = BGC.
Again, ABC can also be written as AGB + AGC + BGC.
$\to$ ABC = 3 AGB or AGB = $\frac{1}{3}$ ABC. Proved.