Show that the diagonals of a square are equal and bisect each other at right angles.

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles.

i.e, AC = BD, OD = OB, OA = OC and \({AC}\perp{BD}\)

Proof: In \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,
BC = AD ...(given)
AB = BA ...(Common side)
\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\)
\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)
\(\therefore \) AC = BD ...(By CPCT)

Similarly, in \(\triangle{OAB}\) and \(\triangle{OCD}\), we have,
AB = DC ...(given)
\(\angle{OAB}\) = \(\angle{DCO}\)
(\(\because \) AB || CD and transversal AC intersect)
\(\angle{OBA}\) = \(\angle{BDC}\)
(\(\because \) AB || CD and transversal BD intersect)
\(\therefore \) \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\)
(By SAS rule)
\(\therefore \) OA = OC and OB = OD
(By CPCT)

Now, in \(\triangle{AOB}\) and \(\triangle{AOD}\), we have,
OA = OD ...(proved earlier)
AB = AD
(sides of square)
AO = OA
(Commom side)
\(\therefore \) \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{AOD}\)
(By SSS rule)
\(\therefore \) \(\angle{AOB}\) = \(\angle{AOD}\) ...(By CPCT)

\(\angle{AOB}\) + \(\angle{AOD}\) = \(180^\circ\)
(linear pair axiom)
\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\)
\(\therefore \) \(AO\perp{BD}\)
\(\Rightarrow \) \(AC\perp{BD}\).
Also, AC = BD, OA = OC, OB = OD and \(AC\perp{BD}\)

Hence, it is proved that diagonals are equal and bisect each other at right angles.