Question

22. Show that the diagonals of a rhombus are perpendicular to
Each other
​

Answer 1

Answer :

ABCD is a rhombus

In which AB = BC = CD = DA

OA = OC AND OB = OD

NOW

IN ▲ BOC AND ▲ DOC

1. OB = OD

2. BC = DC

3. OC = OC

NOW

▲ BOC congruent to  ▲ DOC ( BY SSS)

∠ BOC = ∠ DOC ( BY CPCT)

THEN

∠ BOC +∠ DOC = 180 (LINEAR PAIR)

∠ BOC = ∠ DOC = 90

SIMILARLY

∠ AOB = ∠ AOD = 90

HENCE the diagonals of a rhombus are perpendicular to

Each other

PROVED

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°