Question

Prove that diagonals of square are equal and bisect each other at right angles .

Answer 1

Consider a square ABCD whose diagonals AC and BD intersect each other at a point O.

We have to prove that AC = BD, OA = OC, OB = OD, and ∠AOB = 90º.
Consider ΔABC and ΔDCB,

AB = DC

∠ABC = ∠DCB = 90°

BC = CB (Common side)

∴ ΔABC ≅ ΔDCB (By SAS congruency)

∴ AC = DB (By CPCT)

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

Now, consider ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

AB = CD (Sides of a square are equal)

∴ ΔAOB ≅ ΔCOD (By AAS congruence rule)

∴ AO = CO and OB = OD (By CPCT)

Hence, the diagonals of the square bisect each other.

Similarly, in ΔAOB and ΔCOB,

AO = CO

AB = CB

BO = BO

∴ ΔAOB ≅ ΔCOB (By SSS congruency)

∴ ∠AOB = ∠COB (By CPCT)

∠AOB + ∠COB = 180º (Linear pair)

or, 2∠AOB = 180º

or, ∠AOB = 90º

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

hope this helps

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