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 ≅ ΔEBN
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.