The median BE and CF of a triangle ABC intersect at G. prove that area of triangle-GBC = area of quadrilateral AFGE.
Given: Median of a triangle divides the third side into two equal parts. So, E and F are mid-points of sides AC and AB respectively.
Construction : Join EF.
As we know, the line joining the mid-points of two sides of a triangle is parallel to the third side.
Δ on the same base and between the same parallel lines are equal in area.
∴ ar (BCF) = ar (BCE)
⇒ar (BCG) + ar (CEG) = ar (BCG) + ar (BFG)
⇒ar (CEG) = ar (BFG) ......(i)
The median of a triangle divides the triangle into two triangles of equal area.
BE is median of ΔABC
∴ ar (BCE) = ar (ABE)
⇒ar (BCG) + ar (CEG) = ar (BFG) + ar (AFGE)
⇒ar (BCG) + ar (CEG) = ar (CEG) + ar (AFGE) [From (i)]
⇒ ar (BCG) = ar (AFGE)