In the alongside figure, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that :

(i) AQ = BP

(ii) PQ = CD

(iii) ABPQ is a parallelogram

(i) Let ∠A = 2x.

We know that,

Sum of consecutive angles in a parallelogram equals to 180°.

⇒ ∠A + ∠B = 180°

⇒ 2x + ∠B = 180°

⇒ ∠B = 180° - 2x.

⇒ ∠PAB = $\dfrac{∠A}{2} = \dfrac{2x}{2}$ = x.

⇒ ∠QBA = $\dfrac{∠B}{2} = \dfrac{180° - 2x}{2}$ = 90° - x.

In △ ABP,

⇒ ∠PAB + ∠ABP + ∠BPA = 180° (By angle sum property of triangle)

⇒ x + 180° - 2x + ∠BPA = 180°

⇒ 180° - x + ∠BPA = 180°

⇒ ∠BPA = 180° - 180° + x = x.

∴ ∠BPA = ∠PAB (Both equal to x)

∴ AB = BP (In a triangle sides opposite to equal angles are equal) ……….(1)

In △ ABQ,

⇒ ∠QBA + ∠BAQ + ∠AQB = 180° (By angle sum property of triangle)

⇒ 90° - x + 2x + ∠AQB = 180°

⇒ 90° + x + ∠AQB = 180°

⇒ ∠AQB = 180° - 90° - x = 90° - x.

∴ ∠AQB = ∠QBA (Both equal to 90° - x)

∴ AB = AQ (In a triangle sides opposite to equal angles are equal) ……….(2)

From equation (1) and (2), we get :

⇒ AQ = BP.

Hence, proved that AQ = BP.

(ii) Given,

ABCD is a parallelogram.

We know that,

Opposite sides of parallelogram are equal and parallel.

∴ AB = CD …………(1)

∴ AD || BC.

Since, AD || BC

∴ AQ || BP

Join PQ.

We know that,

AQ = BP (Proved above)

In quadrilateral ABPQ,

AQ = BP and AQ || BP.

Since, one of the pair of opposite sides of quadrilateral ABPQ are equal and parallel.

∴ ABPQ is a parallelogram.

∴ AB = PQ [Opposite sides of parallelogram are equal] ………(2)

From (1) and (2), we get :

PQ = CD.

Hence, proved that PQ = CD.

(iii) In quadrilateral ABPQ,

AQ || BP and AQ = BP.

∴ ABPQ is a parallelogram (Since, one of the pair of opposite sides of quadrilateral ABPQ are equal and parallel.)

Hence, proved that ABPQ is a parallelogram.