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In a triangle ABC, the medians BE and CF intersect at G. Prove that ar (△ BCG) = ar (AFGE).

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Best answer

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).

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