The bisectors of the angles of a parallelogra | vedclass

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(B) In a parallelogram $ABCD$,let the bisectors of $\angle A, \angle B, \angle C,$ and $\angle D$ meet at points $P, Q, R,$ and $S$ as shown in the fi

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Rectangle

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(B) In a parallelogram $ABCD$,let the bisectors of $\angle A, \angle B, \angle C,$ and $\angle D$ meet at points $P, Q, R,$ and $S$ as shown in the figure.Since $ABCD$ is a parallelogram,the sum of adjacent angles is $180^{\circ}$,i.e.,$\angle DAB + \angle ADC = 180^{\circ}$.Since $AS$ and $DS$ are angle bisectors,$\angle DAS = \frac{1}{2} \angle DAB$ and $\angle ADS = \frac{1}{2} \angle ADC$.Therefore,$\angle DAS + \angle ADS = \frac{1}{2} (\angle DAB + \angle ADC) = \frac{1}{2} (180^{\circ}) = 90^{\circ}$.In $	riangle ADS$,by angle sum property,$\angle ASD = 180^{\circ} - 90^{\circ} = 90^{\circ}$.Since vertically opposite angles are equal,$\angle PSR = \angle ASD = 90^{\circ}$.Similarly,it can be shown that $\angle SPQ = 90^{\circ}, \angle PQR = 90^{\circ},$ and $\angle SRQ = 90^{\circ}$.Since all angles are $90^{\circ}$,the quadrilateral $PQRS$ is a rectangle.

The bisectors of the angles of a parallelogram enclose a

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