Question

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

Answer 1

(i) To prove that the diagonals are equal, i.e. AC = BD

In ΔABC and ΔBAD, we have

AB = BA

[Common]

BC = AD

[Opposite sides of the square ABCD]

∠ABC = ∠BAD

[All angles of a square are equal to 90°]

∴ΔABC ≌ ΔBAD

[SAS criteria]

⇒Their corresponding parts are equal.

⇒AC = BD

...(1)

(ii) To prove that 'O' is the mid-point of AC and BD.

∵ AD || BC and AC is a transversal.

[∵ Opposite sides of a square are parallel]

∴∠1 = ∠3

[Interior alternate angles]

Similarly, ∠2 = ∠4

[Interior alternate angles]

Now, in ΔOAD and ΔOCB, we have

AD = CB

[Opposite sides of the square ABCD]

∠1 = ∠3

[Proved]

∠2 = ∠4

[Proved

ΔOAD ≌ ΔOCB

[ASA criteria]

∴Their corresponding parts are equal.

⇒OA = OC and OD = OB

⇒O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O....(2)

(iii) To prove that AC 1 BD.

In ΔOBA and ΔODA, we have

OB = OD

[Proved]

BA =DA

[Opposite sides of the square]

OA = OA

[Common]

∴

ΔOBA ≌ ΔODA

[SSS criteria]

⇒Their corresponding parts are equal.

⇒ ∠AOB = ∠AOD

But ∠AOB and ∠AOD form a linear pair.

∠AOB + ∠AOD = 180°

∠AOB = ∠AOD = 90°

⇒AC ⊥ BD

...(3)

From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.