The problem involves a cyclic quadrilateral ABCD and an equilateral triangle ABC. We need to find the measure of angle CDA.
Since triangle ABC is equilateral, all its angles are $60^{\circ}$. Therefore, $\angle ABC = 60^{\circ}$.
ABCD is a cyclic quadrilateral. A property of cyclic quadrilaterals is that opposite angles sum to $180^{\circ}$. This means $\angle ABC + \angle CDA = 180^{\circ}$.
Substitute the known value of $\angle ABC$ into the equation:
$ 60^{\circ} + \angle CDA = 180^{\circ} $
Solve for $\angle CDA$:
$ \angle CDA = 180^{\circ} - 60^{\circ} $
$ \angle CDA = 120^{\circ} $
Thus, the angle CDA is $120^{\circ}$.
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