Question

In the given figure, ABCD is a trapezium. M and N are mid points of diagonals AC and BD respectively. Which of the following is true?

• A. DC = 2 EB
• B. MN || AB || CD
• C. NE = 0.5 CN
• D. DN = 2 BN

Answer 1

The correct option is B MN || AB || CD

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

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 midpoint 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 midpoints of AC and CE, respectively.

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

$\Rightarrow$ MN || AB || CD