Question

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

Answer 1

According to the question,

It is given that: A square ABCD and OA = OB = AB.

To prove: Δ OCD is an isosceles triangle.

Proof:

In square ABCD,

Since ∠1 and ∠2 is equal to 90^o

∠1 = ∠2 …(1)

Now, in Δ OAB , we have

Since ∠3 and ∠4 is equal to 60^o

∠3 = ∠4 …(2)

Subtracting equations (2) from (1),

We get

∠1−∠3 = ∠2 −∠4

⇒ ∠5 = ∠6

Now,

In Δ DAO and Δ CBO,

AD = BC [Given]

∠5 = ∠6 [Proved above]

OA = OB [Given]

By SAS criterion of congruence,

We have

Δ DAO ≅ Δ CBO

OD = OC

⇒ ΔOCD is an isosceles triangle.

Hence, proved.