The median BE and CF of a triangle ABC intersect at G prove that area of triangle-GBC = area - Maths - Areas of Parallelograms and Triangles

Question

The median BE and CF of a triangle ABC intersect at G prove that
area of triangle-GBC = area of quadrilateral AFGE

Gursheen kaur. · Accepted Answer

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

Proof:

As we know, the line joining the mid-points of two sides of a
triangle is parallel to the third side

âˆ´BC||EF

Î” on the same base and between the same parallel lines
are equal in area

âˆ´ ar (BCF) = ar (BCE)

â‡’ar (BCG) + ar (CEG) = ar (BCG) + ar (BFG)

â‡’ar (CEG) = ar (BFG) (i)

Now,

The median of a triangle divides the triangle into two triangles of
equal area

BE is median of Î”ABC

âˆ´ ar (BCE) = ar (ABE)

â‡’ar (BCG) + ar (CEG) = ar (BFG) + ar
(AFGE)

â‡’ar (BCG) + ar (CEG) = ar (CEG) + ar (AFGE)
[From (i)]

â‡’ ar (BCG) = ar
(AFGE)

The median BE and CF of a triangle ABC intersect at G. prove that area of triangle-GBC = area of quadrilateral AFGE.