Question

show that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus​

Answer 1

suppose ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given,

OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

To show,

ABCD is parallelogram and AB = BC = CD = AD

Proof,

In ΔAOB and ΔCOB,

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.

Thus, AB = BC (by CPCT)

Similarly we can prove,

AB = BC = CD = AD

Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle

Answer 2

Let the quadrilateral ABCD

With diagonals AC and BD intersecting at O

In triangle AOD and triangle BOC

ADO=OBC

DAO=OCB

OD=BO

By AAS theorem triangle AOD is congruent to triangle BOC

in triangle AOB and triangle DOC

OAB=OCD

ABO=ODC

AB=DC

So by SAS theorem triangle AOB is congruent to triangle COD

in triangle AOB and triangle AOD

BO=DO

AOB=AOD

AO=AO

by SAS theorem triangle AOB is congruent to triangle AOD

So,

triangle AOD is congruent to triangle BOC is congruent to triangle AOB is congruent to triangle COD

So,

AB=BC=CD=AD

As in rhombus all sides are equal hence the quadrilateral is rhombus