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.
