Question

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Answer 1

Given: A parallelogram ABCD. The bisectors of interior angles of || gm form a quadrilateral PQRS.

To Prove: PQRS is a rectangle.

Proof:

(1) In || gm ABCD, we have

∠A + ∠B = 180°   ...[Sum of co-interior angles = 180°]

⇒ `1/2 (∠A + ∠B) = 1/2 xx 180^circ`

⇒ `1/2 ∠A + 1/2 ∠B = 90^circ`

∠QAB + ∠QBA = 90°   ...`[{:(∵ (i) AQ  "is bisector of"  ∠A","),(∴ 1/2 ∠A = ∠QAB),((ii) BQ  "is bisector of"  ∠B","),(∴ 1/2 ∠B = ∠QBA):}]`

(2) In ΔAQB, we have

∠QAB + ∠QBA + ∠Q = 180°   ...[Sum of angles of a triangle = 180°]

⇒ 90° + ∠Q = 180°

⇒ ∠Q = 180° – 90°

∠Q = 90°

(3) Similarly, from ΔBCR, we can prove that ∠R = 90°.

From ΔCDS, we can prove that ∠S = 90° and from ΔADP, we can prove that ∠P = 90°

(4) ∠P = ∠Q = ∠R = ∠S = 90°   ...[Proved in (2) and (3)]

Also, ∠P = ∠R and ∠Q = ∠S

i.e. Both pairs of opposite angles are equal.

It is parallelogram with each angle = 90°.

Hence, it is a rectangle.