Question

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

Answer 1

Let ABCD be a square such that its diagonals AC and BD intersect at O.

 

(i) To prove that the diagonals are equal, we need to prove AC = BD.

In ΔABC and ΔBAD, we have

AB = BA           ...[Common]

BC = AD          ...[Sides of a square ABCD]

∠ABC = ∠BAD    ...[Each angle is 90°]

∴ ΔABC ≅ ΔBAD    ...[By SAS congruency]

⇒ AC = BD        ...[By CPCT]          ...(1)

(ii) AD || BC and AC is a transversal.       ...[∵ A square is a parallelogram]

∴ ∠1 = ∠3    ...[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ΔOAD and ΔOCB, we have

AD = CB        ...[Sides of a square ABCD]

∠1 = ∠3        ...[Proved]

∠2 = ∠4       ...[Proved]

∴ ΔOAD ≅ ΔOCB      ...[By ASA congruency]

⇒ OA = OC and OD = OB     ...[By CPCT]       ...(2)

i.e., the diagonals AC and BD bisect each other at O.

(iii) In ΔOBA and ΔODA, we have

OB = OD         ...[Proved]

BA = DA          ...[Sides of a square ABCD]

OA = OA         ...[Common]

∴ ΔOBA ≅ ΔODA        ...[By SSS congruency]

⇒ ∠AOB = ∠AOD        ...[By CPCT]        ...(3)

∵ ∠AOB and ∠AOD form a linear pair

∴ ∠AOB + ∠AOD = 180°

∴ ∠AOB = ∠AOD = 90°        ...[By (3)]

AC ⊥ BD         ...(4)

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.