Let’s prove if on

Given ABCD is a parallelogram in which AB||CD and BC||AD.

∠A=90°

Since, DC||AB,

∠D+∠A=180° (co-interior angles)

⇒ ∠D+90°=180°

⇒ ∠D=180°-90°

⇒ ∠D=90°…………..(1)

Since, AD||BC,

∠B+∠A=180° (co-interior angles)

⇒ ∠B+90°=180°

⇒ ∠B=180°-90°

⇒ ∠B=90°…………..(2)

Also, ∠D+∠C=180° (co-interior angles)

⇒ ∠C+90°=180° (from(1))

⇒ ∠C=180°-90°

⇒ ∠C=90°…………..(3)

Hence, if each angle of a parallelogram is 90° then all the other angles of the parallelogram are 90°.