In the diagram given below, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD. Show that:

(i) ∠PSB + ∠SPB = 90°

(ii) ∠PBS = 90°

(iii) ∠ABC = 90°

(iv) ∠ADC = 90°

(v) ∠A = 90°

(vi) ABCD is a rectangle

(i) Given:

PQRS is a parallelogram.

To prove:

∠PSB + ∠SPB = 90°

Proof:

It is given that AP and CS is an angle bisectors.

∠SPB = ∠BPQ and ∠PSB = ∠BSR

As PS is parallel to QR,

⇒ ∠SPQ + ∠PSR = 180°

⇒ $\dfrac{1}{2}$ (∠SPQ + ∠PSR) = $\dfrac{1}{2}$ 180°

⇒ $\dfrac{1}{2}$ ∠SPQ + $\dfrac{1}{2}$ ∠PSR = 90°

⇒ ∠SPB + ∠PSB= 90°

Hence, ∠PSB + ∠SPB = 90°.

(ii) To prove:

∠PBS = 90°

Proof:

In triangle PBS, the sum of all the angles is 180°.

⇒ ∠PSB + ∠SPB + ∠PBS = 180°

Using ∠PSB + ∠SPB = 90°, we get

⇒ 90° + ∠PBS = 180°

⇒ ∠PBS = 180° - 90°

⇒ ∠PBS = 90°

Hence, ∠PBS = 90°.

(iii)To prove:

∠ABC = 90°

Proof:

∠PBS = ∠ABC (vertically opposite angles)

∠PBS = 90°

So, ∠ABC = 90°

Hence, ∠ABC = 90°.

(iv) To prove:

∠ADC = 90°

Proof:

PQRS is a parallelogram.

As PS is parallel to QR,

⇒ ∠SRQ + ∠RQP = 180°

⇒ $\dfrac{1}{2}$ (∠SRQ + ∠RQP) = $\dfrac{1}{2}$ 180°

⇒ $\dfrac{1}{2}$ ∠SRQ + $\dfrac{1}{2}$ ∠RQP = 90°

⇒ ∠DRQ + ∠RQD = 90°

In triangle RDQ, the sum of all the angles is 180°.

⇒ ∠DRQ + ∠DQR + ∠QDR = 180°

Using ∠DRQ + ∠RQD = 90°, we get

⇒ 90° + ∠QDR = 180°

⇒ ∠QDR = 180° - 90°

⇒ ∠QDR = 90°

∠QDR = ∠ADC (vertically opposite angles)

So, ∠ADC = 90°

Hence, ∠ADC = 90°.

(v) To prove:

∠A = 90°

Proof:

In triangle APQ, the sum of all the angles is 180°.

⇒ ∠APQ + ∠AQP + ∠PAQ = 180°

⇒ (∠APQ + ∠AQP) + ∠PAQ = 180°

[∵ ∠ APQ = $\dfrac{1}{2}$ ∠ SPQ and ∠ AQP = $\dfrac{1}{2}$ ∠ RQP ]

As we know opposite side of parallelogram are equal.

So, ∠ RQP = ∠ PSR

Therefore, ∠ AQP = $\dfrac{1}{2}$ ∠ PSR

⇒ $\dfrac{1}{2}$ (∠SPQ + ∠PSR) + ∠PAQ = 180°

⇒ 90° + ∠PAQ = 180°

⇒ ∠PAQ = 180° - 90°

⇒ ∠PAQ = 90°

Hence, ∠A = 90°.

(vi) To prove:

ABCD is a rectangle.

Proof:

∠A = ∠B = ∠D = 90°

Sum of each angle of quadrilateral is 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 90° + 90° + ∠C + 90° = 360°

⇒ 270° + ∠C = 360°

⇒ ∠C = 360° - 270°

⇒ ∠C = 90°

Each angle of the quadrilateral is 90°.

Hence, ABCD is a rectangle.