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

Solution
Let `E` and `F` are midpoints of the diagonals `AC` and `BD` of trapezium `ABCD` respectively.
Draw `DE` and produce it to meet `AB` at`G`

Consider `/_\AEG` and`/_\CED`

⇒ `/_AEG=/_CED` [ Vertically opposite angles ]

⇒ `AE=EC` [ `E` is midpoint of `AC` ]
⇒ `/_ECD=∠EAG` [ Alternate angles ]
⇒ `/_\AEG≅△CED` [ By SAA congruence rule ]

⇒ `DE=EG` ---- ( 1 ) [ CPCT ]
⇒ `AG=CD` ----- ( 2 )

In `/_\DGB`
`E` is the midpoint of `DG` [ From ( 1 ) ]
`F` is midpoint of `BD`

∴ `EF∥GB`
⇒ `EF∥AB` [ Since `GB` is part of `AB` ]

⇒ `EF` is parallel to `AB` and `CD`.
Also, `EF= 1/2 GB`

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