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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.]
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Solution
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.
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