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

Let ABCD be the parallelogram.

From figure,

AH, BE, CF and DG are bisectors of ∠A, ∠B, ∠C and ∠D respectively.

We know that,

Consecutive angles of parallelogram are supplementary.

Considering ∠A + ∠B = 180°,

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

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

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

Considering ∠B + ∠C = 180°,

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

⇒ $\dfrac{∠B + ∠C}{2}$ = 90°

⇒ $\dfrac{∠B}{2} + \dfrac{∠C}{2}$ = 90° …………(2)

Considering ∠C + ∠D = 180°,

⇒ $\dfrac{∠C + ∠D}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠C + ∠D}{2}$ = 90°

⇒ $\dfrac{∠C}{2} + \dfrac{∠D}{2}$ = 90° ………(3)

Considering ∠D + ∠A = 180°,

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

⇒ $\dfrac{∠D + ∠A}{2}$ = 90°

⇒ $\dfrac{∠D}{2} + \dfrac{∠A}{2}$ = 90° ………(4)

In △ PAB,

By angle sum property of triangle,

⇒ ∠PAB + ∠ABP + ∠BPA = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ + ∠BPA = 180°

⇒ 90° + ∠BPA = 180° [From equation (1)]

⇒ ∠BPA = 180° - 90° = 90°.

From figure,

⇒ ∠SPQ = ∠BPA = 90° (Vertically opposite angles are equal)

In △ BSC,

By angle sum property of triangle,

⇒ ∠SBC + ∠SCB + ∠BSC = 180°

⇒ $\dfrac{∠B}{2} + \dfrac{∠C}{2}$ + ∠BSC = 180°

⇒ 90° + ∠BSC = 180° [From equation (2)]

⇒ ∠BSC = 180° - 90° = 90°.

From figure,

⇒ ∠PSR = ∠BSC = 90°.

In △ DRC,

By angle sum property of triangle,

⇒ ∠RCD + ∠CDR + ∠DRC = 180°

⇒ $\dfrac{∠C}{2} + \dfrac{∠D}{2}$ + ∠DRC = 180°

⇒ 90° + ∠DRC = 180° [From equation (3)]

⇒ ∠DRC = 180° - 90° = 90°.

From figure,

⇒ ∠SRQ = ∠DRC = 90° (Vertically opposite angles are equal)

In △ AQD,

By angle sum property of triangle,

⇒ ∠QAD + ∠ADQ + ∠DQA = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠D}{2}$ + ∠DQA = 180°

⇒ 90° + ∠DQA = 180° [From equation (4)]

⇒ ∠DQA = 180° - 90° = 90°.

From figure,

⇒ ∠RQP = ∠DQA = 90°.

Since, all the interior angles of quadrilateral PQRS equals to 90°.

∴ PQRS is a rectangle.

Hence, proved that bisectors of interior angles of a parallelogram form a rectangle.