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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.

613540_560395_ans_5b056486d14e43e3b2665805002906fb.jpg

Maths

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