Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

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.