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)


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