Question

prove that diagonal of a rhombus are perpendicular to each other

Answer 1

And you see the diagonals intersect at a 90-degree angle. So we've just proved-- so this is interesting. Aparallelogram, the diagonals bisect each other. For arhombus, where all the sides are equal, we've shown that not only do they bisect each other but they'reperpendicular bisectors of each other
draw a figure yourself and then prove it .
mark it the 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°