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The diagonals of a square are equal and bisect each other at right angles.

Detailed proof

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 = 90º.

In ΔABC and ΔDCB,

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

∠ABC = ∠DCB        (All interior angles are of 90)

BC = CB                   (Common side)

So, ΔABC ≅ ΔDCB      (By SAS congruency)

Hence, 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)

So, ΔAOB ≅ ΔCOD  (By AAS congruence rule)

Hence, 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)

So, ΔAOB ≅ ΔCOB       (By SSS congruency)

Hence, ∠AOB = ∠COB      (By CPCT)

However, ∠AOB + ∠COB = 180⁰        (Linear pair)

2∠AOB = 180⁰

∠AOB = 90⁰

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