Question

Question 17
Prove that the line joining the mid-point of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Answer 1

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 $\Delta CDN$ and $\Delta EBN$, we have

DN = BN[since, N is the mid-point of BD]

$\angle DCN=\angle BEN$ [alternate interior angles]

and $\angle CDN=\angle EBN$ [alternate interior angles]

$\therefore$ $\Delta CDN\cong \Delta EBN$ [by AAS congruence rule]

$\therefore$DC = EB and CN =NE[by CPCT rule]

Thus, in $\Delta CAE$, the points M and N are the mid-points of AC and CE, respectively.

$\therefore$MN || AE[by mid-point theroem]

$\Rightarrow$ MN || AB || CD Hence proved.