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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
$$\therefore \angle ABC = \angle BAD...(1) $$ CPCT
Again $$AD$$ $$\parallel$$ $$\mid$$ 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.

490292_463875_ans_80ec7e58be0c4dbeb723c8cffa110ad8.png

Mathematics
RS Agarwal
Standard IX

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