Question

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

Answer 1

CONSIDER RHOMBUS ABCD

YOU KNOW THAT AB=BC=CD=AD

NOW IN Δ AOD AND Δ COD,

OA=OC (DIAGONALS OF A PARELLOGRAM BISECT EACH OTHER)
OD=OD (COMMON)
AD=CD

THEREFORE,Δ AOD CONGRUENT TO Δ COD (SSS)

THIS GIVES ∠ AOD = ∠ COD (CPCT)

BUT, ∠ AOD + ∠ COD = 180 (LINEAR PAIR)

SO, 2 ∠ AOD=180

OR, ∠ AOD =90

SO,THE DIAGONALS OF A RHOMBUS ARE PERPENDICULAR TO EACH OTHER 

HENCE , PROVED

Hope This Helps :)

Answer 2

Answer:

Given:

▶ABCD is a rhombus

.•.AB = BC = CD = DA

To prove:

▶<AOB = <BOC = <COD = <DOA

Proof:

▶In ∆AOD and ∆COD,

AO = OC [diagonal of a parallelogram bisect each other]

OD = OD [common]

AD = CD [given]

▶Therefore,

∆AOD ≈ ∆COD [SSS congruence rule]

.•.<AOD = <COD [CPCT]

▶But,

<AOD + <COD = 180° [linear pair]

▶so,

2<AOD = 180°

▶or,

<AOD = 90°

Hope it's helpful ✅✅✅