Question

In a triangle ABC,the medians BE and CF intersect at G.Prove that $ar\left(△BCG\right)=ar\left(AFGE\right)$.

Answer 1

ANSWER:
Figure
CF is median of △ABC.
⇒ar(△BCF) = 1/2(△ABC)
.....(1)
Similarly, BE is the median of △ABC,
⇒ar(△ABE) = 1/2(△ABC)
.....(2)
From (1) and (2) we have
ar(△BCF) = ar(△ABE)
⇒ar(△BCF) - ar(△BFG) = ar(△ABE) - ar(△BFG)
⇒ar(∆BCG) = ar(AFGE)