Question

show that the diagonals of a square are equal and perpendicular to each other

Answer 1

Sol:

Given that ABCD is a square.

To prove : AC = BD and AC and BD bisect each other at right angles.

Proof: 

(i)  In a Δ ABC and Δ BAD,

AB =  AB ( common line)

BC = AD ( opppsite sides of a square)

∠ABC = ∠BAD ( = 90° )

Δ ABC ≅ Δ BAD ( By SAS property)

AC = BD ( by CPCT).

(ii) In a Δ OAD and Δ OCB,

AD = CB ( opposite sides of a square)

∠OAD = ∠OCB ( transversal AC )

∠ODA = ∠OBC ( transversal BD )

ΔOAD ≅ ΔOCB (ASA property)

OA = OC  ---------(i)

Similarly OB = OD ----------(ii)

From (i) and (ii)  AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB = OD ( from (ii) )

BA = DA

OA = OA  ( common line )

ΔAOB = ΔAOD ----(iii) ( by CPCT

∠AOB + ∠AOD = 180°   (linear pair)

2∠AOB  = 180°

∠AOB = ∠AOD = 90°

∴AC and BD 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