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Answer :

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.

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