  Question

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

Solution

Given: In parallelogram $$ABCD$$, $$AC = BD$$To prove : Parallelogram $$ABCD$$  is rectangle.Proof : in $$\triangle ACB$$ and $$\triangle BDA$$$$AC = BD$$  $$\mid$$ Given$$AB = BA$$ $$\mid$$ Common$$BC = AD$$ $$\mid$$ Opposite sides of the parallelogram ABCD$$\triangle ACB$$ $$\cong \triangle BDA \mid SSS$$ Rule$$\therefore \angle ABC = \angle BAD...(1)$$ CPCTAgain $$AD$$ $$\parallel$$ $$\mid$$ Opposite sides of parallelogram $$ABCD$$$$AD$$ $$\parallel BC$$ and the traversal $$AB$$ intersects them.$$\therefore \angle BAD + \angle ABC = 180^{\circ}$$ ...$$(2)$$ _ Sum of consecutive interior angles on the same side of the transversal is $$180^{\circ}$$From (1) and (2) ,$$\angle BAD = \angle ABC = 90^{\circ}$$$$\therefore \angle A = 90^{\circ}$$ and $$\angle C = 90^{\circ}$$Parallelogram $$ABCD$$ is a rectangle. MathematicsRS AgarwalStandard IX

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