Question

in the alongside diagram, the bisectors of
interior angles of the parallelogram PQRS
enclose a quadrilateral ABCD.
how that:
(0) ZPSB + ZSPB = 90° (ii) ZPBS = 90°
(iii) ZABC = 90° (iv) ZADC = 90°
(v) ZA = 90° (vi) ABCD is a rectangle
Thus, the bisectors of the angles of a
parallelogram enclose a rectangle.
​

Answer 1

in the alongside diagram, the bisectors of  interior angles of the parallelogram PQRS  enclose a quadrilateral ABCD.

• In parallelogram PQRS,
• PS || QR (opposite sides)
• ∠P + ∠Q = 180°
• AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram
• ∠APQ + ∠AQP = 1/2 × 180° = 90°
• But in ∆APQ,
• ∠A + ∠APQ + ∠AQP = 180° (Angles of a triangle)
• ∠A + 90° = 180°
• ∠A = 180° – 90°
• (v) ∠A = 90°
• Similarly PQ || SR
• (i) ∠PSB + ∠SPB = 90°
• (ii)  ∠PBS = 90°
• But, ∠ABC = ∠PBS (Vertically opposite angles)
• (iii) ∠ABC = 90°
• Similarly
• (iv) ∠ADC = 90° and ∠C = 90°
• (vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)
• Hence proved.