Let ABCD be a trapezium in which AB||DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

Now, join CN and produce it to AB at E.

In ΔCDN and ΔEBN, we have,

DN = BN

∠DCN = ∠BEN

∠CDN = ∠EBN

ΔCDN EBN

DC = EB and CN = NE

Thus, in ΔCAE, the points M and N are the mid-points of AC and CE, respectively.

MN||AE

MN||AB||CD

Hence, proved.