91Ó°ÊÓ

It is being given that ABCD is a parallelogram.

In a parallelogram, both pairs of opposite sides are congruent and parallel.

Therefore, in the parallelogram ABCD, both pairs of opposite sides are congruent and parallel.

Therefore, $\stackrel{¯}{AB}â‰\dots \stackrel{¯}{CD}$, $\stackrel{¯}{\mathrm{AD}}â‰\dots \stackrel{¯}{\mathrm{BC}}$, $\stackrel{¯}{\mathrm{AB}}âˆ\yen \stackrel{¯}{\mathrm{CD}}$and $\stackrel{¯}{\mathrm{AD}}âˆ\yen \stackrel{¯}{\mathrm{BC}}$.

Therefore, $\mathrm{AB}=\mathrm{CD}$.

It is also being given that M and N are the midpoints of $\stackrel{¯}{\mathrm{AB}}$and $\stackrel{¯}{\mathrm{DC}}$.

As, M is the midpoint of $\stackrel{¯}{\mathrm{AB}}$, therefore by using the definition of midpoint it can be said that role="math" localid="1637741738802" $\mathrm{AM}=\frac{1}{2}\mathrm{AB}$.

As, N is the midpoint of $\stackrel{¯}{\mathrm{DC}}$, therefore by using the definition of midpoint it can be said that role="math" localid="1637741802557" $\mathrm{NC}=\frac{1}{2}\mathrm{DC}$.

Therefore, it can be noticed that:

$\begin{array}{c}\mathrm{AB}=\mathrm{CD}\\ \frac{1}{2}\mathbf{AB}\mathbf{=}\frac{1}{2}\mathbf{DC}\\ \mathrm{AM}=\mathrm{NC}\end{array}$

Therefore, $\mathbf{AM}â‰\dots \mathbf{NC}$.

As, $\stackrel{¯}{\mathrm{AB}}âˆ\yen \stackrel{¯}{\mathrm{CD}}$, therefore it can be said that $\stackrel{¯}{\mathrm{AM}}âˆ\yen \stackrel{¯}{\mathrm{NC}}$.

If one pair of opposite side is both congruent and parallel, then the quadrilateral is a parallelogram.

As, $\stackrel{¯}{\mathrm{AM}}âˆ\yen \stackrel{¯}{\mathrm{NC}}$and $\mathbf{AM}â‰\dots \mathbf{NC}$, therefore, the quadrilateral AMCN is a parallelogram.