Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of trapezium and is half of their difference.

Explanation

Let `E` and `F` are midpoints of the diagonals `AC` and `BD` of trapezium `ABCD` respectively.
we have to Draw `DE` and produce it to meet `AB` at `G`
Consider `△AEG` and `△CED`
`=>∠AEG=∠CED `
`=>AE=EC `
`=>∠ECD=∠EAG `
`=>△AEG≅△CED`
`=>DE=EG`
`=>AG=CD`
In `△DGB`

`E` is the midpoint of `DG`
`F` is midpoint of `BD`

`∴ EF∥GB `
`=>EF∥AB `
`=> EF` is parallel to `AB and CD`.
Also, `EF=1/2GB`
`=>EF=1/2(AB-AG)`
`=>EF=1/2(AB-CD)`