Question

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆OCD is an isosceles triangle.

Answer 1

Given: O is a point in the interior of a square ABCD such that ΔOAB is an equilateral triangle.

Construction: Join OC and OD.

To show: ΔOCD is an isosceles triangle.

Proof: Since, AOB is an equilateral triangle.

∴ ∠OAB = ∠OBA = 60°   ...(i)

Also, ∠DAB = ∠CBA = 90°  ...(ii) [Each angle of a square is 90°] [∵ ABCD is a square]

On subtracting equation (i) from equation (ii), we get

∠DAB – ∠OAB = ∠CBA – ∠OBA = 90° – 60°

i.e. ∠DAO = ∠CBO = 30°

In ΔAOD and ΔBOC,

AO = BO  ...[Given] [All the side of an equilateral triangle are equal]

∠DAO = ∠CBO   ...[Proved above]

And AD = BC   ...[Sides of a square are equal]

∴  ΔAOD ≅ ΔBOC  ...[By SAS congruence rule]

Hence, OD = OC  ...[By CPCT]

In ΔCOD,

OC = OD

Hence, ΔCOD is an isosceles triangle.

Hence proved.