Show that the segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two equal parallelograms.

Explanation

GIVEN A ||gm ABCD in which E and F are the midpoints of AB and DC respectively. E and F are joined.
TO PROVE `ar(squareAEFD)=ar(squareEBCF).`
PROOF Since ABCD is a ||gm, we have AB||DC
and AB = DC.
`therefore AE"||"DFand AE=DF`
`[therefore AB=DCrArr(1)/(2)DCrArrAE = DF]`
`therefore` AEFD is a ||gm.
Again, EB||FC and EB = FC
`[therefore AB=DC rArr(1)/(2)AB=(1)/(2)DCrArr EB=FC]`
`therefore` EBCF is a ||gm.
Now, AEFD and EBCF being two parallelograms with equal bases and between the same parallels, they must have equal areas.
`therefore ar("||gm AEFD") = ar("|| gm EBCF").`