Question

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Answer 1

Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of $\angle A,\angle B,\angle Cand\angle D$, respectively.

To prove Quadrlateral PQRS is a rectangle.

Proof Since, ABCD is a parallelogram, then DC||AB and DA is a transversal.
We have, $\angle A+\angle D={180}^{\circ }$
[sum of cointerior angles of a parallelogram is ${180}^{\circ }$]
$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle D={90}^{\circ }$ [dividing both sides by 2]
$\Rightarrow \angle PAD+\angle PDA={90}^{\circ }$
$\Rightarrow \angle APD={90}^{\circ }$ [since, sum of all angles of a triangle is ${180}^{\circ }$]
$\therefore \angle SPQ={90}^{\circ }$ [vertically opposite angles]
Similarly, $\angle PQR={90}^{\circ }$
$\angle QRS={90}^{\circ }$
$PSR={90}^{\circ }$

Thus, PQRS is a quadrilateral whose each angle is ${90}^{\circ }$.

Hence, PQRS is a rectangle.