Question

$\text{In the given figure, O is a point in the interior of square ABCD such that}\Delta OAB\text{is an equilateral triangle.}\text{Show that}\Delta OCD\text{is an isosceles triangle.}$

Answer 1

Given: In square ABCD, Δ OAB is an equilateral triangle.
To prove: Δ OCD is an isosceles triangle.
Proof:
∵∠DAB=∠CBA=90°

⇒∠OAD=∠OBC=30°

Angles of square ABCD And, ∠OAB= ∠OBA=60° .....i [Angles of equilateral ∆OAB]

∴∠DAB-∠OAB=∠CBA-∠OBA=90°-60=30°\

Now, in Δ DAO and Δ CBO,

AD = BC (Sides of square ABCD)
∠ DAO = ∠ CBO [From (i)]
AO = BO (Sides of equilateral Δ OAB)

∴ By SAS congruence criteria,

Δ DAO ≅ Δ CBO

So, OD = OC (CPCT)

Hence, Δ OCD is an isosceles triangle.