According to the question,
We have,
BE & CF are medians
E is the midpoint of AC
F is the midpoint of AB
∴ ΔBCE = ΔBEA … ( i )
ΔBCF = ΔCAF
Construct:
Join EF,
By midpoint theorem,
We get FE || BC
We know that,
Δ on the same base and between same parallels are equal in area
∴ ΔFBC = ΔBCE
ΔFBC – ΔGBC = ΔBCE – ΔGBC
⇒ ΔFBG = ΔCGE (ΔGBC is common)
⇒ ΔCGE = ΔFBG …( ii )
Subtracting equation (ii) from (i)
We get,
ΔBCE – ΔCGE = ΔBEA – ΔFBG
∴ ΔBGC = quadrilateral AFGE.
