**Given :** ABCD is a parallelogram in which AC = BD.

**To Prove :** ABCD is a rectangle.

**Proof :** In ∆ABC and ∆ABD

AB = AB [Common]

BC = AD [Opposite sides of a parallelogram]

AC = BD [Given]

∴ ∆ABC ≅ ∆BAD [SSS congruence]

∠ABC = ∠BAD …(i) [CPCT]

Since, ABCD is a parallelogram, thus,

∠ABC + ∠BAD = 180° …(ii) [Consecutive interior angles]

∠ABC + ∠ABC = 180°

∴ 2∠ABC = 180° [From (i) and (ii)]

⇒ ∠ABC = ∠BAD = 90°

This shows that ABCD is a parallelogram one of whose angle is 90°. Hence, ABCD is a rectangle. **Proved.**