2. In a cyclic quadrilateral $A B C D$, it holds that $A B=3, B C=6$ and $\triangle A C D$ is equilateral. Let $O$ be the center of the circumscribed circle around quadrilateral $A B C D$, and $E$ the intersection of diagonals $A C$ and $B D$. Calculate $\measuredangle D O E$.
2. Since $\measuredangle A B D$ and $\measuredangle C B D$ are both $60^{\circ}$ (due to the equality of peripheral angles), we conclude that $B D$ is the angle bisector of $\measuredangle A B C$, and thus $\frac{A E}{E C}=\frac{A B}{B C}=\frac{1}{2}$. Let $F$ be the midpoint of segment $A C$. Then from the above, we have $\frac{A E}{E F}=2$, and since $\frac{D O}{O F}=2$ (because $\triangle A C D$ is equilateral), by the converse of Thales' theorem, we have $E O \| A D$. From this, it follows that $\measuredangle F O E=\measuredangle F D A=30^{\circ}$, and then $\measuredangle D O E=150^{\circ}$.