jmc

First, let us draw $CO$: Since $AO=CO,$ we have that $△AOC$ is isosceles, and so $\angle CAO=\angle ACO=18^{\circ }.$ We are given that $AC=BC,$ we can see that $△BOC\cong △AOC,$ so $\angle BCO=18^{\circ }.$ Now, $\angle ACB=\angle OAC+\angle BCO=36^{\circ },$ and as $\angle ACB$ is an inscribed angle, we have $\angle AOB=2\cdot \angle ACB=\boxed{72^{\circ }}.$