sir, please answer this question......

Construction: Join EF

Since the line segment joining the mid points of two sides of a triangle is parallel to the third side.

We have, EF || BC

Triangles BEF and CEF are the same base EF and between the same parallel lines.

So, ar(BEF) = ar(CEF)

i.e ar(BEF) - ar(GEF) = ar(CEF) - ar (GEF)

ar (BFG) = (CEG)

We know that the median of a triangle divides it into two triangles of equal area.

So, ar(BEC) = ar(ABE)

i.e ar(BGC) + ar(CEG)= ar(quad.AFGE) + ar(BFG)

i.e ar(BGC) + ar(BFG)= ar(quad.AFGE) + ar(BFG)

Therefore, i.e ar(BGC) = ar(quad.AFGE)