If bisectors of ∠A and ∠B of a parallelogram ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a :

1. rectangle

2. rhombus

3. parallelogram

4. quadrilateral whose opposite angles are supplementary

∠A + ∠D = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\dfrac{1}{2}∠A + \dfrac{1}{2}∠D$ = 90°

In triangle ASD,

∠DAS + ∠SDA + ∠ASD = 180°

$\dfrac{1}{2}∠A + \dfrac{1}{2}∠D$ + ∠ASD = 180°

90° + ∠ASD = 180°

∠ASD = 90°

∠PSR = ∠ASD = 90° [Vertically opposite angles]

∠S = 90°

∠B + ∠C = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\dfrac{1}{2}∠B + \dfrac{1}{2}∠C$ = 90°

In triangle BQC,

∠QCB + ∠QBC + ∠CQB = 180°

$\dfrac{1}{2}∠C + \dfrac{1}{2}∠B$ + ∠CQB = 180°

90° + ∠CQB = 180°

∠CQB = 90°

∠PQR = ∠CQB = 90° [Vertically opposite angles]

∠Q = 90°

∠A + ∠B = 180° [sum of Co-Int. angles in ∥gm is 180°]

⇒ $\dfrac{1}{2}∠A + \dfrac{1}{2}∠B$= 90°

In triangle APB,

$\dfrac{1}{2}∠A + \dfrac{1}{2}∠B$ + ∠APB = 180°

90° + ∠APB = 180°

∠APB = 90°

∠P = 90°

In quadrilateral PQRS,

By angle sum property of quadrilateral:

∠P + ∠Q + ∠R + ∠S = 360°

90° + 90° + ∠R + 90° = 360°

∠R + 270° = 360°

∠R = 360° - 270°

∠R = 90°

Since, all angles of PQRS = 90°

∴ PQRS is a rectangle

Hence, option 1 is the correct option.