A $\Delta ABC$ has vertices lying on a circle $w$ of radius $1$, with $\angle BAC = 60^\circ$. A circle with center $I$ is inscribed in $\Delta ABC$. 
A $\Delta ABC$ has vertices lying on a circle $w$ of radius $1$, with $\angle BAC = 60^\circ$. A circle with center $I$ is inscribed in $\Delta ABC$. The line $AI$ meets circle $w$ at $D$. Find $ID$.

What I Tried: I am assuming the circle with center $I$ is the incircle of $\Delta ABC$. So here is a picture :-

Let $O$ be the center of the circumcircle. Then all the red-marked angles are $30^\circ$ each, and the green angles are $120^\circ$ each. So, simply from properties of cyclic quadrilaterals, one can find that $\square BOCD$ is a rhombus  with :- $BO = OC = OD = CD = DB$ .
But that does not seem to help on how to find $ID$ , I thinking I am missing something here which I could use. Also after checking this through Geogebra I find $ID$ to be miraculously $1$ cm, the same length of the radius of the circumcircle.
Can anyone help me how are we finding this?

Hint. Prove that $ID=DB=DC$ using angle-chasing. (This lemma is actually known as Fact 5 in the Olympiad context). You should then have $ID=DB=R=1$, since $\triangle OBD$ is equilateral.

 Proof of the Lemma "Fact 5".
Consider the triangle $\triangle IDB$ and notice that $$\angle BID=\angle BAI+\angle IBA=\frac{\alpha+\beta}2$$ At the same time, observe that $$\angle DBI=\angle DBC+\angle CBI=\angle DAC+\angle CBI=\frac{\alpha+\beta}2$$ Thus, $\triangle IBD$ is isosceles with $ID=DB$.

