Question

18. The diagonals AC and BD of a rhombus intersect each other at O. Prove that:
AB²+ BC²+CD + DA²= 4(0A²+OB²)
(Hint: The diagonals of a rhombus bisect each other at right angles.)​

Answer 1

Step-by-step explanation:

Given :-

• The diagonals AC and BD of a rhombus intersect each other at ‘O’.

To prove :-

• AB²+ BC² + CD² + DA² = 4(OA² +OB²)

Solution :-

AC and BD are the diagonals of the rhombus. AC and BD intersect each other at the point ‘O’.

• AO = OC
• BO = OD

The diagonals of a rhombus bisect each other at right angles. (Given)

★In case of ∆ AOB,

AB² = OA² + OB²...............(i)

★In case of ∆ BOC,

BC² = OB² + OC² ...............(ii)

★ In case of ∆ COD,

CD² = OC² + OD²................(iii)

★ In case of ∆ DOA,

DA² = OA² + OD² .................(iv)

Now,

(i)+(ii)+(iii)+(iv)

AB² + BC² + CD² + DA² = OA² + OB² + OB² + OC² + OC² + OD² + OA² + OD²

→ AB² + BC² + CD² + DA² = 2OA² + 2OB² + 2OC² + 2OD²

→ AB²+ BC²+ CD²+ DA² = 2OA² + 2OB² + 2OA² + 2OB² [ •°• OA = OC , OB = OD]

→ AB²+ BC²+ CD²+ DA² = 4OA² + 4OB²

→ AB²+ BC²+ CD²+ DA² = 4(OA² + OB²)

Hence proved !

Answer 2

Answer:

It is easy friend

Step-by-step explanation:

As all the triangles formed by the diagonals of the rhombus are Right angled triangle.

taking️ triangle ABO ,️ triangle BOC,️triangle DOC and️ triangle DAO.

To prove:-AB²+BC²+CD²+DA²=4(OA²+OB²)

proof:-

In triangle ABO

Using pythagoras theorem,

AB²=OB²+AO² ....(i)

in triangle BOC

Using pythagoras theorem,

CB²=OB²+OC² ....(ii)

in triangle DAO

using pythagoras theorem,

AD²=DO²+OA²....(iii)

in triangle DOC

using pythagoras theorem,

CD²=DO²+OC²....(iv)

Adding equations (i),(ii),(iii),(iv).

we get :-

AB²+BC²+CD²+DA²=OB²+AO²+OB²+OC²+DO²+OC²+DO²+AO²

or, AB²+BC²+CD²+DA²= 2(OB²+AO²+OC²+DO²)

We know,

OB=DO(given)

or,OB²=DO²(Taking DO² as OB² as both are equal)

and, OC=AO(given)

or,OC²=AO²(Taking OC² as AO² as both are equal)

so,

AB²+BC²+CD²+DA²= 2(OB²+OC²+OC²+OB²)

or,AB²+BC²+CD²+DA²=4(OB²+OC²)

Hence proved,

*Thank Uhhhh!!!*