Question

1. Prove that the diagonals of a square are equal and perpendicular to each other

Answer 1

We know that the diagonals of a rectangle are equal.
Also, we know that every square is a rectangle.
So, the diagonals of a square are equal.
Again, we know that the diagonals of a rhombus bisect each other at right angles.
But every square is a rhombus.
So, the diagonals of a square bisect each other at right angles.
Hence, the diagonals of a square are equal and bisect each other at right angles.

Answer 2

Given :- ABCD is a square.

To proof :- AC = BD and AC ⊥ BD

Proof :- In △ ADB and △ BCA

AD = BC [ Sides of a square are equal ]

∠BAD = ∠ABC [ 90° each ]

AB = BA [ Common side ]

△ADB ≅ △BCA [ SAS congruency rule ]

⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]

In △AOB and △AOD

OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]

AB = AD [ Sides of a square are equal ]

AO = AO [ Common side ]

△AOB ≅ △ AOD [ SSS congruency rule ]

⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]

∠AOB + ∠AOD = 180° [ Linear pair ]

∠ AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD

Hence proved, AC = BD and AC ⊥BD