Question

Solve this question :-

Show that the diagonal of a rhombus are perpendicular to each other.

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Answer 1

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Given :- ABCD is a rhombus

AC and BD are diagonals of rhombus intersecting at O.

⇒ To prove :- ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°

⇒ Proof :- All Rhombus are parallelogram, Since all of its sides are equal.

AB = BC = CD = DA ────(1)

The diagonal of a parallelogram bisect each other

Therefore, OB = OD and OA = OC ────(2)

In ∆ BOC and ∆ DOC

BO = OD [ From 2 ]

BC = DC [ From 1 ]

OC = OC [ Common side ]

∆ BOC ≅ ∆ DOC [ By SS congruency criteria ]

∠BOC = ∠DOC [ C.P.C.T ]

∠BOC + ∠DOC = 180° [ Linear pair ]

2∠BOC = 180° [ ∠BOC = ∠DOC ]

∠BOC = 180°/2

∠BOC = 90°

∠BOC = ∠DOC = 90°

Similarly, ∠AOB = ∠AOD = 90°

Hence, ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°

Answer 2

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Consider the rhombus ABCD

You know that AB = BC = CD = DA (Why ?)

Now, in △AOD and △COD,

OA = OC ( Diagonal of a parallelogram bisect each other )

OD = OD ( Common )

AD = CD

Therefore, △AOD ≅ △COD ( SSS congruence rule)

This gives, ∠AOD + ∠COD. ( CPCT )

But, ∠AOD + ∠COD = 180°. ( Linear pair )

So, 2 ∠AOD = 180°

Or, ∠AOD = 90°

So, the diagonal of a rhombus are perpendicular to each other.

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