Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Given: A quadrilateral ABCD in which AC = BD and AC \(\perp{BD}\) such that OA = OC and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.
In \(\triangle{ABO}\) and \(\triangle{ADO}\), we have,
BO = OD ...(given)
AO = OA ...(Common side)
\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\) ...(given)
\(\therefore \) \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ADO}\)
(By SAS rule)
\(\therefore \) AB = AD ...(By CPCT)
Also, AB = DC and AD = BC
(opposite sides of parallelogram)
\(\therefore \) AB = BC = DC = AD ...(i)

Similarly, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,
AC = BD ...(given)
AB = BA ...(Common side)
BC = AD ...(from(i))
\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)
(By SSS rule)
\(\therefore \) \(\angle{ABC}\) = \(\angle{BAD}\) ...(ii)(By CPCT)

But \(\angle{ABC}\) + \(\angle{BAD}\) = \(180^\circ\) >br> (linear pair axiom)
\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\) ...(from (ii))
\(\therefore \) AB = BC = CD = DA, and \(\angle{A}\) = \(90^\circ\)

Hence, it is proved that, ABCD is a square.