Question

prove that diagonal of a rhombus are perpendicular to each other

Answer 1

Let ABCD is a rhombus and O is the point of intersection of the diagonals

so AB=BC=CD=AD
diagonals also bisect each other, so OA=OC and OB=OD

now in triangle BOC and DOC
OB=OD, BC=CD and OC=OC

so triangle BOC is congruent to triangle DOC by SSS
so by cpct, angle BOC = angle DOC

but angle BOC + angle DOC = 180° (LP)
so angle BOC = angle DOC = 90°

similarly angle AOB = angle AOD = 90°

Hence, the diagonals of a rhombus are perpendicular to each other

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°