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).

Explanation

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 `DeltaBDA` into two triangles of equal area.
`therefore" "` `ar (DeltaBRA) = (1)/(2) ar (DeltaBDA)" "` ...(i)
Similarly, Median RS divides `DeltaBRA` into two triangles of equal area.
`therefore" "` `ar (DeltaASR) = (1)/(2) ar (DeltaBRA)" "` ...(ii)
From Eqs. (i) and (ii),
`ar (DeltaASR) = (1)/(4) ar (DeltaBDA)" "` ...(iii)
Similarly, `" " ar (DeltaCFP) = (1)/(4) ar (DeltaBCD)" "` ...(iv)
On adding Eqs. (iii) and (iv), we get
`ar (DeltaASR) + ar (DeltaCFP) = (1)/(4) ar (DeltaBDA)" "` [`ar (DeltaBDA) + ar (DeltaBCD)`]
`rArr" "` `ar (DeltaASR) + ar (DeltaCFP) = (1)/(4)ar ("quadrilateral" BCDA)" "` ...(v)
Similarly, `" " ar(DeltaDRF) + ar (DeltaBSP) = (1)/(4)ar ("quadrilateral" BCDA)" "` ...(vi)
On adding Eqs. (v) and (vi), we get
`ar(DeltaASR) + ar(DeltaCFP) + ar (DeltaDRF) + ar (DeltaBSP)`
`=(1)/(2)ar ("quadrilateral" BCDA)" "` ...(vii)
But `ar (DeltaASR) + ar (DeltaCFP) + ar (DeltaDRF) + ar (DeltaBSP) + ar ("parallelogram" PFRS)`
`= ar ("quadrilateral" BCDA)" "` ...(viii)
On subtracting Eq. (vii) from Eq. (viii), we get
`ar ("parallelogram" PFRS) = (1)/(2)ar ("quadrilateral" BCDA)" "` Hence proved.