Question

In the following diagram, 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
Thus, the bisectors of the angles of a parallelogram enclose a rectangle.

Answer 1

Given: In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.

To prove :
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 9°
(vi) ABCD is a rectangle

Proof : In parallelogram PQRS,

PS || QR (opposite sides)

(i) ∠P +∠Q = 180°

= `(∠P)/2 + (∠S)/2 = 90°`

= ∠SPB + ∠PSB = 90°`

(ii) In tringle ∠PBS = 90°

∠SPB + ∠PSB + ∠PBS = 180°

= 90° + ∠PSB = 180°

= ∠PSB = 90°

(iii) ∠ABC = 90°

∠PBS = ∠ABC {vertically opposite}

∠ABC = 90°

(iv) In ∠ADC = 90° 

`R/2 + Q/2 + RDQ = 180°`

RDQ = 180 - 90

RDQ = 90° 

∴ ∠ADC = 90°  {vertically opposite}

(v) ∠A = 90°

Similarly PQ || SR

∠PSB + SPB = 90°

(vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)

Hence proved.