Question

In which shape are the diagonals equal in length and bisect each other at 90°?

Answer 1

The diagonals in a square bisect each other at 90° and are equal in length.



To prove:

AC = BD

AO = OC

and ∠AOB = 90°

In ΔABC and ΔBAD,

AB = BA (Common)

∠ABC = ∠BAD = 90°

BC = AD (Given)

ΔABC ≅ ΔBAD [SAS congruency]

Hence,

AC = BD [CPCT]

diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

ΔAOB ≅ ΔCOD [AAS congruency]

Hence,

AO = CO [CPCT].

Diagonals bisect each other.

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

ΔAOB ≅ ΔCOB [SSS congruency]

also, ∠AOB = ∠COB

∠AOB+∠COB = 180° (Linear pair)

Hence,

∠AOB = ∠COB = 90°

Diagonals bisect each other at right angles

Hence, proved.