In the diagram below; P and Q are midpoints of sides BC and AD respectively of the parallelogram ABCD. If side AB = diagonal BD; prove that the quadrilateral BPDQ is a rectangle.

Given: ABCD is a parallelogram. P and Q are midpoints of sides BC and AD, respectively.

AB = BD

In a parallelogram, opposite sides are equal and parallel. Thus, AB = CD and AD = BC.

Since P and Q are midpoints, by the Midpoint Theorem,

PQ || AB and PQ = $\dfrac{1}{2}$ AB.

Similarly, BP = QD.

In Δ QBD and Δ DPB,

BD = BD (Common)

QD = BP (midpoint property, QD = $\dfrac{1}{2}$ AD = $\dfrac{1}{2}$ BC = BP)

∠PBD = ∠QDB (alternate interior angles, since PQ || AB)

Using SAS congruency criterion,

Δ QBD ≅ Δ DPB

Therefore, by corresponding parts of congruent triangles,

QB = DP

In quadrilateral BPDQ,

QD = BP

QB = DP

Thus, opposite sides of BPDQ are equal.

In Δ QAB and Δ DQB,

AQ = QD (Q is mid-point of AD)

AB = BD (Given)

QB = QB (Common side)

Using SSS congruency criterion,

Δ QAB ≅ Δ DQB

Therefore, by corresponding parts of congruent triangles,

∠AQB = ∠DQB

⇒ ∠AQB + ∠DQB = 180° (Linear pair of angles)

⇒ ∠AQB + ∠AQB = 180°

⇒ 2∠AQB = 180°

⇒ ∠AQB = $\dfrac{180°}{2}$

⇒ ∠AQB = ∠DQB = 90°

Thus, BPDQ has opposite sides equal and two adjacent angles 90°. Since one angle of a parallelogram is 90°, all angles are 90°.

Since BPDQ has opposite sides equal and all angles 90°, it is a rectangle.

Hence, the quadrilateral BPDQ is a rectangle.