Question

If the diagonals of a parallelogram are equal, then show that it is a rectangle.
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Answer 1

$\sf \underline{ \dag \: Solution \: : }$

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Given : in parallelogram ABCD, AC = BD

To Prove : parallelogram ABCD is a rectangle.

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Proof :

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In ∆ACB and ∆BDA

AC = BD | Given

AB = AB | Common

BC = AD | Opposite sides of the parallelogram

∆ACB ≅ ∆BDA

∴ ∠ABC = ∠BCD (C.P.C.T) ...i)

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Again,

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AD || BC | Opposite sides of the parallelogram

AD || BC and the transversal AB intersects them

∴ ∠BAD + ∠ABC = 180° | sum of the consecutive interior angles on the same side of the transversal is 180° ...ii)

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From i) and ii)

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∠BAD = ∠ABC = 90°

∴ ∠A = 90° ∠C = 90°

So, parallelogram ABCD is a rectangle.

Answer 2

Answer:

Given : in parallelogram ABCD, AC = BD

To Prove : parallelogram ABCD is a rectangle.

Proof :

In ∆ACB and ∆BDA

AC = BD | Given

AB = AB | Common

BC = AD | Opposite sides of the parallelogram

∆ACB ≅ ∆BDA

∴ ∠ABC = ∠BCD (C.P.C.T) ...i)

Again,

AD || BC | Opposite sides of the parallelogram

AD || BC and the transversal AB intersects them

∴ ∠BAD + ∠ABC = 180° | sum of the consecutive interior angles on the same side of the transversal is 180° ...ii)

From i) and ii)

∠BAD = ∠ABC = 90°

∴ ∠A = 90° ∠C = 90°

So, parallelogram ABCD is a rectangle.