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

O is a point in the interior of square ABCD. ∆OAB is an equilateral triangle.

Now,

∠DAB = ∠CBA .....(1) (Measure of each angle of a square is 90º)

∠OAB = ∠OBA .....(2) (Measure of each angle of an equilateral triangle is 60º)

Subtracting (2) from (1), we get

∠DAB − ∠OAB = ∠CBA − ∠OBA

⇒ ∠OAD = ∠OBC

In ∆OAD and ∆OBC,

OA = OB (Sides of an equilateral triangle are equal)

∠OAD = ∠OBC (Proved above)

AD = BC (Sides of a square are equal)

∴ ∆OAD ≅ ∆OBC (SAS congruence axiom)

⇒ OD = OC (CPCT)

In ∆OCD,

OC = OD

∴ ∆OCD is an isosceles triangle. (A triangle whose two sides are equal is an isosceles triangle)