Question

good morning to everyone Show that the diagonals of a rhombus are perpendicular to each other?

Answer 1

In Rhombus ABCD, AC and BD are diagonals which meet at O.

In triangles ABO and CBO,

AB=BC -Sides of a rhombus

angleBAO=angleBCO -- Angles opposite to equal sides are equal

AO=CO --diagonals bisect each other

Therefore, Triangle ABO congruent to CBO

angleAOB=angleBOC -- CPCT

but they are linear pair. so,

AOB+BOC=180o

AOB+AOB=180o

2AOB=180o

AOB=BOC=90o

DOA=COB=90o --V.O.A

DOC=AOB=90o -- V.O.A

DOA=DOC=COB=AOB=90o

HENCE 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°