Question

show that diagonals of a rhombus are perpendicular to each other

Answer 1

Answer:

Step-by-step explanation:

CONSIDER RHOMBUS ABCD

YOU KNOW THAT AB=BC=CD=AD

NOW IN TRIANGLE AOD AND TRIANGLE COD

OA=OC(DIAGONALS OF A //GM BISECT EACH OTHER)

OD=OD (COMMON)

AD=CD

THEREFORE,TRIANGLE AOD CONGRUENT TO TRIANGLE COD (SSS)

THIS GIVES ANGLE AOD = ANGLE COD (CPCT)

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

SO, 2 ANGLE AOD=180

OR, ANGLE AOD =90

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

HENCE , PROVED

Please mark as brainliest.

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°