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

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

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Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To show that,

AC = BD

AO = OC

and ∠AOB = 90°

Proof:

In ΔABC and ΔBAD,

AB = BA (Common)

∠ABC = ∠BAD = 90°

BC = AD (Given)

ΔABC ≅ ΔBAD [SAS congruency]

AC = BD [CPCT]

diagonals are equal.

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

ΔAOB ≅ ΔCOD [AAS congruency]

Thus,

AO = CO [CPCT].

Diagonal bisect each other.

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

ΔAOB ≅ ΔCOB [SSS congruency]

∠AOB = ∠COB

∠AOB + ∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90°

Diagonals bisect each other at right angles

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