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Question
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 || DC and let M and N be the mid-points of the diagonals AC and BD, respectively.
To prove: MN || AB || CD
Construction: Join CN and produce it to meet AB at E.
In ΔCDN and ΔEBN, we have
DN = BN ...[Since, N is the mid-point of BD]
∠DCN = ∠BEN ...[Alternate interior angles]
And ∠CDN = ∠EBN ...[Alternate interior angles]
∴ ΔCDN ≅ ΔEBN ...[By AAS congruence rule]
∴ DC = EB and CN = NE ...[By CPCT rule]
Thus, in ΔCAE, the points M and N are the mid-points of AC and CE, respectively.
∴ MN || AE ...[By mid-point theorem]
⇒ MN || AB || CD
Hence proved.
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