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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.