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


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


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