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

(N/A) Given: $A$ square $ABCD$ and $\triangle OAB$ is an equilateral triangle,so $OA = OB = AB$.
To prove: $\triangle OCD$ is an isosceles triangle.
Proof: In square $ABCD$,$\angle DAB = \angle CBA = 90^{\circ}$.
Since $\triangle OAB$ is equilateral,$\angle OAB = \angle OBA = 60^{\circ}$.
Now,$\angle DAO = \angle DAB - \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Similarly,$\angle CBO = \angle CBA - \angle OBA = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle DAO$ and $\triangle CBO$:
$AD = BC$ (Sides of a square)
$\angle DAO = \angle CBO = 30^{\circ}$ (Proved above)
$OA = OB$ (Sides of an equilateral triangle)
By $SAS$ congruence criterion,$\triangle DAO \cong \triangle CBO$.
Therefore,$OD = OC$ (Corresponding parts of congruent triangles).
Since $OD = OC$,$\triangle OCD$ is an isosceles triangle.
Hence,proved.