Let ABCD be a parallelogram. To show that ABCD is a rectangle,

We have to prove that: One of its interior angles is 90⁰

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

ΔABC ΔDCB (By SSS Congruence rule)

∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180⁰

∠ABC + ∠DCB = 180⁰ (AB || CD)

∠ABC + ∠ABC = 180⁰

∠ABC = 180⁰

2 ∠ABC = 90⁰

Since ABCD is a parallelogram and one of its interior angles is 90⁰, ABCD is a rectangle