If AD is a median

Construction: Draw AM⊥ BC

Proof: Since,

AD is the median of ΔABC

Therefore,

BD = DC

BD * AM = DC * AM

(BD * AM) = (DC * AM)

Area (Δ ABD) = Area (Δ ACD) (i)

Now, in Δ BGC

GD is the median

Therefore,

Area (BGD) = Area (CGD) (ii)

Also,

In Δ ACD, CG is the median

Therefore, Area (AGC) = Area (CGD) (iii)

From (i), (ii) and (iii) we have

Area (ΔBGD) = Area (ΔAGC)

But,

Area (ΔBGC) = 2 Area (ΔBGD)

Therefore,

Area (BGC) = 2 Area (ΔAGC)

Hence, proved