Question 9
If the midpoints 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).

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Solution

Let ABCD be a quadrilateral and P, F, R and S are the midpoints of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.

To prove:
ar (parallelogram PFRS) = $\frac{1}{2}$ ar (quadrilateral ABCD)

Construction: Join BD and BR.

Proof:
Median BR divides $Δ$ BDA into two triangles of equal area.
$a r (Δ B R A) = \frac{1}{2} a r (Δ B D A) . . . (i)$

Similarly, median RS divides $Δ$ BRA into two triangles of equal area .
$a r (Δ A S R) = \frac{1}{2} a r (Δ B R A) . . . (i i)$

From eqs. (i) and (ii);
$a r (Δ A S R) = \frac{1}{4} a r (Δ B D A) . . . (i i i)$

Similarly, $a r (Δ C F P) = \frac{1}{4} a r (Δ B C D) . . . . (i v)$

On adding eqs. (iii) and (iv), we get;
$a r (Δ A S R) + a r (Δ C F P) = \frac{1}{4} [a r (Δ B D A) + a r (Δ B C D)]$
$\Rightarrow a r (Δ A S R) + a r (Δ C F P) = \frac{1}{4} a r (q u a d r i l a t e r a l B C D A) . . . (v)$

Similarly, $a r (Δ D R F) + a r (Δ B S P) = \frac{1}{4} a r (q u a d r i l a t e r a l B C D A) . . . . (v i)$

On adding equations (v) and (vi), we get,
$a r (Δ A S R) + a r (Δ C F P) + a r (Δ D R F) + a r (Δ B S P)$
$= \frac{1}{2} a r (q u a d r i l a t e r a l B C D A) . . . (v i i)$

But,
$a r (Δ A S R) + a r (Δ C F P) + a r (Δ D R F) + a r (Δ B S P) + a r (p a r a l l e l o g r a m P F R S)$
$= a r (q u a d r i a l a t e r a l B C D A) . . . . (v i i i)$

On subtracting Eq. (vii) from EQ (viii), we get,
$a r (∥ g m P F R S) = \frac{1}{2} a r (Q u a d r i l a t e r a l B C D A)$

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