Question

The angle-bisectors of a parallelogram forms a quadrilateral as shown in the figure. Show that ABCD is a rectangle.

Answer 1

Given quadrilateral PQRS is a paralleogram.
$\therefore$ $\angle$QPS + $\angle$PSR = ${180}^{\circ }$ (because consecutive angles of a parallelogram are supplementary) (1)

As, PC and SA are the respective angle bisectors of $\angle$P and $\angle$S,
on dividing both sides of (1) by 2, we have,
$\angle$DPS + $\angle$PSD = ${90}^{\circ }$

Thus, in $△$PDS, $\angle$PDS = ${90}^{\circ }$
or, $\angle$ADC = $\angle$PDS = ${90}^{\circ }$ (vertically opposite angles)
Similarly, $\angle$DCB = $\angle$CBA = $\angle$BAD = ${90}^{\circ }$
Thus, fig.ABCD is a rectangle.