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

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Solution.

Given: Let ABCD be a trapezium in which AB\parallel DC  and let M and N be the mid-points of diagonals AC and BD.

To Prove: MN\parallel AB\parallel CD
Proof: Join CN and produce it to meet AB at E
In \triangle CDN and \triangle EBN we have
DN = BN {N is mid-point of BD}
\angle DCN=\angle BEN {alternate interior angle}
\angle CDN=\angle EBN {alternate interior angles}
\triangle CDN\cong \triangle EBN 
DC = EB and CN = NE {by CPCT}
Thus in \triangle CAE, the points M and N are the mid-points of AC and CE, respectively.

MN\parallel AE   {By mid-point theorem}
MN\parallel AB\parallel CD

Hence Proved

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