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  • 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 (Fig.).

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 (Fig.).

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Solution.
Given that if the mid-points of the sides of a quadrilateral are joined in order, a parallelogram is formed. We have to find the area of this parallelogram.
To prove: ar (parallelogram PFRS) =\frac{1}{2} ar(quadrilateral ABCD)

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.
Construction:- Join BD and BR.
We know that median of a triangle divides it into two triangles of equal area.
So, BR divides \triangle BDA into two triangles of equal area.
\therefore ar(\triangle BRA) =\frac{1}{2} ar(\triangle BDA)                    …(i)
Similarly, median RS divides \triangle BRA into two triangles of equal area.
\therefore ar(\triangle ASR)=\frac{1}{2}ar(\triangle BRA)                     …(ii)
From eq. (i) and (ii) 
\therefore ar(\triangle ASR)=\frac{1}{4}ar(\triangle BDA)                    …(iii)
Similarly,
ar(\triangle CFP)=\frac{1}{4}ar(\triangle BCD)                       ….(iv)
On adding equations (iii) and (iv), we get
ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar(\triangle BDA+\triangle BCD)
\Rightarrow ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar(Quadrilateral BCDA) …(v)
Similarly,
ar(\triangle DRF)+ar(\triangle BSP)=\frac{1}{4}ar(BCDA)….(vi)
On adding (v) and (vi), we get

ar(\triangle ASR)+ar(\triangle CFP)+ar (\triangle BSP)+ar(\triangle BSP)    
=\frac{1}{2}ar (quadrilateral BCDA)                                                          …(vii)
But
ar(\triangle ASR) + ar(\triangle CFP) + ar(\triangle DRF) + ar(\triangle BSP) = ar(\parallel gm PFRS)
=\frac{1}{2}ar       (quadrilateral BCDA)                                                               …(viii)
from subtracting eq. (vii) from eq. (viii) we get
ar(parallelogram PFRS)  =\frac{1}{2}ar (quadrilateral BCDA)

Hence proved

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