Question

# Prove that if the diagonals of a parallelogram are equal then it is a rectangle.

Solution

## Let $$ABCD$$ be a parallelogram. To show that $$ABCD$$ is a rectangle, we have to prove that one of its interior angles is $$90^0$$.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 $$180^0$$.$$∠ABC + ∠DCB = 180^0$$ (AB || CD)$$⇒ ∠ABC + ∠ABC = 180^0$$$$⇒ 2∠ABC = 180^0$$$$⇒ ∠ABC = 90^0$$Since $$ABCD$$ is a parallelogram and one of its interior angles is $$90^0$$.Hence, $$ABCD$$ is a rectangle.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More