Solution:

Let $O$ be the center of $\omega$, and let the two tangent lines intersect at $P$. Note that $O$ lies on the external angle bisector of $\angle CPE$ because the tangents are symmetric about line $PO$. Additionally, $O$ lies on the perpendicular bisector of $CE$ by symmetry. By Fact 5, $COPE$ is cyclic and $\angle COE = 90^{\circ}$. To finish, observe that $\angle COD = 45^{\circ}$. Dropping the altitude $CH$ down to $AD$ gives $OH = CH = \sqrt{3}$. So, $AO = AH - OH = 3 - \sqrt{3}$. The desired answer is then $\frac{\sqrt{3}}{2} \cdot AO = \left[\frac{3\sqrt{3} - 3}{2}\right]$.

Solution 2: Another way to get $\angle COE = 90^{\circ}$ is as follows.
Let $\omega$ meet the tangents from $C$ and $E$ at $Q$ and $R$, respectively. Observe $OC = OE$ (as $O$ lies on the perpendicular bisector of $CE$) and $OQ = OR$, so $\triangle OCQ \cong \triangle OER$. Then $\angle COE = \angle QOR = 90^{\circ}$.