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

ABCD is a //gm.

From figure,

∠A + ∠D = 180° [Sum of Co-Interior angles in a // gm is 180°]

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

In triangle APD,

∠DAP + ∠PDA + ∠APD = 180°

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

90° + ∠APD = 180°

∠APD = 90°

∠SPQ = ∠APD = 90° [Vertically opposite angles are equal]

∠P = 90°

From figure,

∠B + ∠C = 180° [Sum of Co-Interior angles in a // gm is 180°]

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

In triangle BRC,

∠CBR + ∠BCR + ∠CRB = 180°

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

90° + ∠CRB = 180°

∠CRB = 90°

∠SRQ = ∠CRB = 90° [Vertically opposite angles are equal]

∠R = 90°

From figure,

∠A + ∠B = 180° [Sum of Co-Interior angles in a // gm is 180°]

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

In triangle ASB,

∠SAB + ∠SBA + ∠ASB = 180°

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

90° + ∠ASB = 180°

∠ASB = 90°

∠S = 90°

In quadrilateral PQRS,

By angle sum property of quadrilateral,

∠P + ∠Q + ∠R + ∠S = 360°

90° + ∠Q + 90° + 90° = 360°

∠Q + 270° = 360°

∠Q = 360° - 270° = 90°.

Since, all the angles of PQRS = 90°

∴ PQRS is a rectangle.

Hence, proved that PQRS is a rectangle.