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Answer :

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

\(\Rightarrow \) OA = OC and OB = OD

And also,

\(\angle{AOD} = \angle{AOB} = \angle{COD} = \angle{BOC} = 90^\circ\).

To prove: ABCD is a rhombus.

Proof:

In \(\triangle{OAB}\) and \(\triangle{ODC}\), we have,

OA = OC and OB = OD ...(given)

\(\angle{AOB}\) = \(\angle{COD}\)

(vertically opposite angles)

\(\therefore \) \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\)

(By SAS rule)

\(\therefore \) AB = CD ...(i)(By CPCT)

Similarly, in \(\triangle{OAD}\) and \(\triangle{OBC}\), we have,

OA = OC and OD = OB ...(given)

\(\angle{AOD}\) = \(\angle{BOC}\)

(vertically opposite angles)

\(\therefore \) \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCB}\) ...(By SAS rule)

\(\therefore \) AD = BC ...(ii)(By CPCT)

Similarly, we can prove that,

AB = CD and

CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,

AB = BC = AD = CD

Hence, it is proved that ABCD is a rhombus.

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