jmc

Because $\overline{AD}=\overline{CD}$, $△ADC$ is an isosceles triangle with $\angle DAC=\angle DCA$. Angles in a triangle add up to $180^{\circ }$, and since $\angle ADC=60^{\circ }$, the other two angles are also $60^{\circ }$, and $△ADC$ is an equilateral triangle. Therefore $\overline{AC}=\overline{DA}=\boxed{17}$.