Draw a line EF.
We know that the line segment joining the midpoint of two sides of a triangle is parallel to the third side.
So we get
FE || BC
From the figure we know that △ BEF and △ CEF lie on the same base EF between the same parallel lines
So we get
Area of △ BEF = Area of △ CEF
By subtracting △ GEF both the sides
Area of △ BEF – Area of △ GEF = Area of △ CEF – Area of △ GEF
We get
Area of △ BFG = Area of △ CEG …… (1)
Median of a triangle divides it into two triangles having equal area
Area of △ BEC = Area of △ ABE
It can be written as
Area of △ BGC + Area of △ CEG = Area of quadrilateral AFGE + Area of Area of △ BFG
Using the equation (1) we get
Area of △ BGC + Area of △ BFG = Area of quadrilateral AFGE + Area of Area of △ BFG
We get
Area of △ BGC = Area of quadrilateral AFGE
Therefore, it is proved that ar (△ BCG) = ar (AFGE).