Question

Show that the bisectors of the angles of a parallelogram enclose a rectangle.​

Answer 1

$\huge{\underline{\underline{\sf{\pink{Solution}}}}}$

Since DC || AB and DA cuts them,we have

$\angle \: A + \:\angle \: D \: = \: 180\degree \: \: \: \: (co - interior \: \angle \:s)$

$⇒ \frac{1}{2} \:\angle \: A \: + \frac{1}{2} \angle \: D = 90\degree$

$⇒ \angle \: PAD \: + \angle \: PDA \: = 90\degree$

$⇒ \angle \: APD = 90\degree \: \: \: \: \: (\therefore \: sum \: of \: the \: \angle \: s \: of \: a \: \triangle \: is \:180 \degree )$

$⇒ \angle \: SQP \: =90 \degree \: \: \: \: \: \: \: \: (vert. \: opp. \: \angle \: s).$

Now,AD || BC and DC cuts them.

$\therefore \: \angle \: D \: + \: \angle \: C \: =180 \degree \: \: \: \: \: \: (co - interior \: \angle \: s)$

$⇒ \: \frac{1}{2} \angle \: D \: + \frac{1}{2} \angle \: C = 90\degree \:$

$⇒ \angle \: QDC + \angle \: QCD \: =90 \degree$

$⇒ \angle \: DQC =90 \degree \: \: \: \: \: [\therefore \: sum \: of \: the \: \angle \: s \: of \: a \: \triangle \: is \: 180\degree]$

$⇒ \angle \: PQR \: =90 \degree.$

$similarly \: \angle \: QRS = 90\degree \: and \: \angle \: RSP = 90\degree$

Thus,PQRS is a quadrilateral beach of whosr angles is 90°.

Hence,PQRS is a rectangle.