If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Given: ABCD is a parallelogram with AC = BD.
To Prove: ABCD is a rectangle.
Solution: In ΔABC and Δ BAD,
BC = AD [∵ Opposite sides of a parallelogram are equal]
AC = BD [∵ Given]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule]
∠ABC = ∠BAD [∵ CPCT]
But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]
⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]
⇒ ∠BAD = 180°/2= 90°
A parallelogram with one of its is 90° is a rectangle. Hence, ABCD is a rectangle.