91Ó°ÊÓ

The opposite sides of a parallelogram are congruent.

In $â–\pm QRST$, $\stackrel{¯}{QR}$and $\stackrel{¯}{TS}$are opposite sides. Therefore, $\stackrel{¯}{QR}â‰\dots \stackrel{¯}{TS}$.

From the figure, it can be observed that $\stackrel{¯}{QR}âˆ\yen \stackrel{¯}{TS}$and $\stackrel{¯}{TR}$is a transversal then $∠STM$and $∠MRQ$are alternate interior angles such that, $∠STMâ‰\dots ∠MRQ$.

From the figure, it can be observed that $\stackrel{¯}{QR}âˆ\yen \stackrel{¯}{TS}$and $\stackrel{¯}{SQ}$is a transversal then $∠TSM$and $∠MQR$are alternate interior angles such that, $∠TSMâ‰\dots ∠MQR$.

As $∠STMâ‰\dots ∠MRQ$, $\stackrel{¯}{QR}â‰\dots \stackrel{¯}{TS}$and $∠TSMâ‰\dots ∠MQR$then by ASA postulate $\mathrm{Δ}QMRâ‰\dots \mathrm{Δ}SMT$.

As $∠TSMâ‰\dots ∠MQR$then by corresponding parts of congruent triangles, $\stackrel{¯}{QM}â‰\dots \stackrel{¯}{MS}$and $\stackrel{¯}{RM}â‰\dots \stackrel{¯}{MT}$.

It is given that $\stackrel{¯}{QS}$and $\stackrel{¯}{TR}$are the diagonals of a parallelogram. From $\stackrel{¯}{QM}â‰\dots \stackrel{¯}{MS}$and $\stackrel{¯}{RM}â‰\dots \stackrel{¯}{MT}$ it can be concluded that the diagonals of a parallelogram bisect each other at point M.

Hence it is proved that diagonals of a parallelogram bisect each other.