Question

show that the diagonals of a rhombus are perpendicular to each other

Answer 1

Heya!

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Diagnol of a rhombus are perpendicular to each other.

Given - Rhombus ABCD, AC and BD are bisectors intersecting at O

To prove - AC ⊥ BD

Proof - In ∆AOB and ∆COD

OA = OC (Diagnol of llgm bisect each other)
OD = OD (Common)
AD = CD

Thus, ∆AOD ≅ ∆COD (SSS Congruence)

Also, ∠AOD = ∠COD (C.P.C.T)

But, ∠AOD + ∠COD=180° (Linear pair)
Thus, 2 ∠AOD= 180
∠AOD = 180/2 = 90°

Thus, Diagnol of a rhombus is perpendicular to each other.

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Hope it helps...!!!

Answer 2

⇒ 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°