Question

DC, if MN = 15 cm and AB = 23 cm.

Answer 1

Join BD, let BD and MN meet at Q. Since, M is the mid point of AD and N is the mid point of BC. So by mid point theorem, AB II MN IICD
In $△$, BDC and BQN,
$\angle B=\angle B$ (Common)
$\angle BDC=\angle BQN$ (Corresponding angles of parallel lines)
$\angle BCD=\angle BNQ$ (Corresponding angles of parallel lines)
thus, $△BDC\sim △BQN$
Thus, $\frac{BD}{QB}=\frac{DC}{QN}$
$2=\frac{DC}{QN}$ (Q is the mid point of BD)
$QN=\frac{1}{2}DC$
Similarly, $QM=\frac{1}{2}AB$
Hence, $QM+QN=\frac{1}{2}(AB+DC)$
$MN=\frac{1}{2}(AB+CD)$
Hence, $2\times 15=23+CD$
$CD=30-23=7cm$