Question

show that if the diagonal of a quadrilateral bisect each other at right angles, then is a rhombus.​

Answer 1

$\boxed{\mathcal{\blue{ANSWER}}}$

To prove -:

If diagonals of a quadrilateral bisect each other ar right angles, then it is a rhombus.

Proof-:

Let a quadrilateral ABCD whose diagonals intersect at O.

In ∆AOB and ∆AOD,

OB = OD (Given)

AO = AO (Common)

∠AOB = ∠AOD (90°)

∆AOB ≅ ∆AOD (by SAS criteria)

∴AB = AD (by c.p.c.t)............(i)

Now,

In ∆BOC and ∆COD,

OB = OD (given)

CO = CO (common)

∠BOC = ∠COD(90°)

∆BOC ≅ ∆COD (by SAS criteria)

∴BC = CD (by c.p.c.t).............(ii)

Similarly,

We can prove that,

AB = BC..........(iii)

CD = AD..........(iv)

From (i),(ii),(iii) and (iv)

AB = BC = CD = AD

Since, all the sides of a rhombus are equal and it is given that the diagonals bisect at 90°,then ABCD is a rhombus.

Answer 2

Ello there!

Let ABCD be a quadrilateral such that its diagonals AC and BD bisect each other at right angles. Then,

OA = OC, OB = OD, ∠AOB = ∠BOC = 90° and ∠AOD = ∠COD = 90°

Since diagonals AC and BD of quadrilateral ABCD bisect each other. Therefore, ABCD is a parallelogram.

In △s AOB and BOC, we have

OA = OC

∠AOB = ∠COB = 90°

and, OB = OB

=> △AOB ≅ △BOC [ SAS Rule ]

Also, AB = CB (c.p.c.t.)

Since ABCD is a parallelogram.

Hence, AB = DC and BC = AD.

So, AB = BC = CD = AD

Hence, ABCD is a rhombus. _____[PROVED]