Given the diagram and that AC = AO, show that C2 = D1.

Since O is the center of the circle, OA and OC are radii. Therefore, $OA = OC$.
Given $AC = AO$, it follows that $AC = AO = OC$.
Thus, $\triangle AOC$ is an equilateral triangle.
Therefore, $\angle OAC = \angle OCA = \angle AOC = 60^\circ$.
From the diagram, $\angle OCA$ is $\hat{C}_1$, so $\hat{C}_1 = 60^\circ$.