Question

the median BE and CF of a triangle ABC intersect at G. prove that area(triangle GBC) is equal to area(quadrilateral AFGE)

Answer 1

Given that 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.

Proof:

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

∴BC||EF

Triangle 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)

Now, 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)

$|proved|$