Question

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).

[Hint: Join BD and draw perpendicular from A on BD.]

Answer 1

Given: Let ABCD is a quadrilateral and P, F, R and S are the mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.

To prove: ar (parallelogram PFRS) = `1/2` ar (quadrilateral ABCD)

Construction: Join BD and BR.

Proof: Median BR divides ΔBDA into two triangles of equal area.

∴ ar (ΔBRA) = `1/2` ar (ΔBDA)  ...(i)

Similarly, median RS divides ΔBRA into two triangles of equal area.

∴ ar (ΔASR) = `1/2` ar (ΔBRA)  ...(ii)

From equations (i) and (ii),

ar (ΔASR) = `1/4` ar (ΔBDA)  ...(iii)

Similarly, ar (ΔCFP) = `1/4` ar (ΔBCD)  ...(iv)

On adding equations (iii) and (iv), we get

ar (ΔASR) + ar (ΔCFP) = `1/4` ar (ΔBDA)  ...[ar (ΔBDA) + ar (ΔBCD)]

⇒ ar (ΔASR) + ar (ΔCFP) = `1/4` ar (quadrilateral BCDA)  ...(v)

Similarly, ar (ΔDRF) + ar (ΔBSP) = `1/4` ar (quadrilateral BCDA)  ...(vi)

On adding equations (v) and (vi), we get

ar (ΔASR) + ar (ΔCFP) + ar (ΔDRF) + ar (ΔBSP) = `1/2` ar (quadrilateral BCDA)   ...(vii)

But ar (ΔASR) + ar (ΔCFP) + ar (ΔDRF) + ar (ΔBSP) + ar (parallelogram PFRS) = ar (quadrialateral BCDA)  ...(viii)

On subtracting equation (vii) from equation (viii), we get

ar (parallelogram PFRS) = `1/2` ar (quadrilateral BCDA)

Hence proved.