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: A quadrilateral ABCD in which the mid-points of the sides of it are joined in order to form parallelogram PQRS.
TO prove: `ar(||gm PQRS) = (1)/(2) ar (square ABCD)`
Consturction : Join BD and draw perpendicular from A on BD which intersect SR and BD at X and Y respectively.
Proof: In `Delta ABD`, S and R are the mid-points of sides AB and AD respectively

`:. SR || BD`
`rArr SX||BY`
`rArr` X is the mid-point of AY (converse of mid-point theorem)
`rArr AX = XY` ...(1)
and `SR = (1)/(2) BD` ( `:'` mid-point theorem)..(2)
Now, `ar(Delta ABCD) = (1)/(2) xx BD xx AY`
and `ar(Delta ASR) = (1)/(2) xx SR xx AX`
`rArr ar (Delta ASR) = (1)/(2) xx ((1)/(2)BD) xx ((1)/(2) AY)` [using (1) and (2)]
`rArr ar(Delta ASR) = (1)/(4) xx ((1)/(2) BD xx AY)`
`rArr ar (Delta ASR) = (1)/(4) ar (Delta ABD)`...(3)
Similarly,
`ar(Delta CPQ) = (1)/(4) ar (Delta CBD)`...(4)
`ar(Delta BPS) = (1)/(4) ar(Delta BAC)`...(5)
`ar (Delta DRQ) = (1)/(4) ar (Delta DAC)`...(6)
Adding eqs. (3), (4), (5) and (6), we get
`ar(Delta ASR) + ar(Delta CPQ) + ar (Delta BPS) + ar (Delta DRQ)`
`= (1)/(4) ar (Delta ABD) + (1)/(4) ar (Delta CBD) + (1)/(4) ar(Delta BAC) + (1)/(4) ar(Delta DAC)`
`= (1)/(4) [ar (Delta ABD) + ar(Delta CBD) + ar (Delta BAC) + ar (Delta DAC)]`
`= (1)/(4) [ar (square ABCD) + ar(squareABCD)] = (1)/(4) xx 2 ar (square ABCD) = (1)/(2) ar (square ABCD)`
`:. ar (Delta ASR) + ar (Delta CPQ) + ar (Delta BPS) + ar(Delta DRQ) = (1)/(2) ar (square ABCD)`
`rArr ar(square ABCD) - ar (||gm PQRS) = (1)/(2) ar (square ABCD)`
`rArr ar (||gm PQRS) = ar(square ABCD) - (1)/(2) ar (squareABCD)`
rArr ar (||gm PQRS) = (1)/(2) ar (square ABCD)`