Let P, Q, R, S be respectively the midpoints of the sides AB, BC, CD and DA of quad. ABCD Show that PQRS is a parallelogram such that
`ar("||gm PQRS")=(1)/(2)ar("quad. ABCD")`.

Explanation

Join AC and AR.
In `triangleABC`, P and Q are midpoints of AB and BC respectively.
`therefore` PQ||AC and PQ `=(1)/(2) AC`.
In `triangleDAC`, S and R are midpoints of AD and DC respectively.
`therefore` SR||AC SR`=(1)/(2) AC`.
Thus, PQ||SR and PQ = SR.
`therefore` PQRS is a ||gm.
Now, median AR divides `triangle`ACD into two `triangle` of equal area.
`therefore ar(triangleARD)=(1)/(2)ar(triangleACD)." "...(i)`
Median RS divides `triangle`ARD into two `triangle` of equal area.
`therefore ar(triangleDSR)=(1)/(2)ar(triangleARD)." "...(ii)`
From (i) and (ii), we get `ar(triangleDSR)=(1)/(4)ar(triangleACD)`.
Similarly, `ar(triangleBQP)=(1)/(4)ar(triangleABC)`.
`therefore ar(triangleDSR)+ar(triangleBQP)=(1)/(4)[ar(triangleACD)+ar(triangleABC)]`
`rArrar(triangle DSR)+ar(triangleBQP)=(1)/(4)ar["quad.ABCD"]." "...(iii)`
Similarly, `ar(triangleCRQ)+ar(triangleASP)=(1)/(4)ar("quad. ABCD")." "...(iv)`
Adding (iii) and (iv), we get
`ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)`
`=(1)/(2)ar("quad. ABCD")" "...(v)`
But, `ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)`
`+ar("||gm PQRS)=ar("quad. ABCD")." "...(vi)`
Substracting (v) from (vi), we get
`ar("||gm PQRS") =(1)/(2)ar("quad. ABCD")`.