Answer :

**Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.**

**To prove that the diagonals of a square are equal and bisect each other at right angles,**

We have to prove: AC = BD, OA = OC, OB = OD and ∠AOB = 90^{0}

In ΔABC and ΔDCB,

AB = DC (Sides of a square are equal to each other)

∠ABC = ∠DCB (angles of square are 90°)

BC = CB (Common side)

ΔABC ΔDCB (By SAS congruency)

AC = DB (By CPCT)

Hence, the diagonals of a square are equal in length.

In ΔAOB and ΔCOD,

∠AOB =∠ COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal)

ΔAOB ΔCOD (By AAS congruence rule)

AO = CO and

OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other

In ΔAOB and ΔCOB,

As we had proved that diagonals bisect each other, therefore,

AO = CO

AB = CB (Sides of a square are equal)

BO = BO (Common)

ΔAOB ΔCOB (By SSS congruency)

∠AOB =∠ COB (By CPCT)

However,

∠ AOB + ∠COB = 180^{0} (Linear pair)

2∠AOB = 180^{0}

∠AOB = 90^{0}

Hence, the diagonals of a square bisect each other at right angles

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