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

Explanation

Given: In trapezium, `ABCD`, `AB` parallel `CD` and `P`, `Q` are the midpoints of `AC` and `BD`.
We need to prove `PQ` parallel `AB` and `DC`, `PQ=1/2(AB-DC)`
Construction: Join `DP` and produce `DP` to meet `AB` in `R`.
Proof: Since `AB` parallel `DC` and transversal `AC` cuts them.
`/_1=/_2` …..(i) [Alternate interiror angles are equal]
Now, in `/_\APR` and `/_\DPC`, we have:
`/_1=/_2` [From (i)]
`AP=CP` [`P` is the mid-point of `AC`]
And, `/_3=/_4` [Vertically opposite angles]
`=>/_\APR=/_\CPD` [By `ASA`]
`=>AR=DC` and `PR=DP [CPCT]` …….(ii)
In `/_\DRB`,
`P` and `Q` are the mid points of sides `DR` and `DB` respectively
`=>PQ` parallel `RB`
`=>PQ` parallel `AB` [`RB` is a part of `AB`]
`PQ` parallel `AB` and `PQ` parallel `DC` `[AB || DC]` (Given)
Again,` P` and `Q` are the mid- point of sides `DR` and `DB` respectively in `/_\DRB`.
`PQ=1/2RB`
`=>PQ=1/2(AB–AR)`
`=>PQ=1/2(AB–DC)` [From (ii), `AR=DC`]
Hence Proved.