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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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