prove that if the diagonals of a parallelogram are equal,then its a rectangle Share with your friends Share 18 Divyangana Bhaskar answered this Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.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 = ∠DCBIt is known that the sum of the measures of angles on the same side of transversal is 180º.∠ABC + ∠DCB = 180º (AB || CD)⇒ ∠ABC + ∠ABC = 180º⇒ 2∠ABC = 180º⇒ ∠ABC = 90ºSince ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle. 63 View Full Answer Srivatsan C answered this LET THE QUADRILATERAL BE ABCD .AC = BD .AB = ABAD = BCSO , TRIANGLE ABD IS CONGRUENT TO TRIANGLE BACSO , ANGLE A IS EQUAL TO ANGLE B ( BY CPCT )BUT ANGLE A + ANGLE B = 180SO , ANGLE A IS 90ANGLE B = 90AS ANGLE A + ANGLE = 180ANGLE D = 180ANGLE A = ANGLE CANGLE C = 90 2