BE is the median of ΔABC and we know that a median of a triangle divides it into two parts of equal area.
⇒ar(ΔABE)=ar(ΔCBE)
ar(ΔABE)=12×ar(ΔABC) ...(i)
Similarly, ar(ΔACF)=ar(ΔBCF)
⇒ar(ΔBCF)=12×ar(ΔABC) ...(ii)
From eqs. (i) and (ii),
ar(ΔABE)=ar(ΔBCF) ....(iii)
On subtracting ar(ΔGBF) from both sides of Eq. (iii), we get,
⇒ar(ΔABE)−ar(ΔGBF)=ar(ΔBCF)−ar(ΔGBF)
⇒ar(AFGE)=ar(GBC)