Let $ \{X\} = OC_1\cap h_c$, where $ h_c$ is the altitude corresponding to the side $ AB$. Moreover, let $ K$ be the intersection point of the line $ CC_1$ with the circumcenter of $ \triangle{ABC}$. Consider the inversion $ I$ of pole $ O$ and power $ CO^2$. Then $ I(C_1) = X$, thus $ COKX$ is a cyclic trapezoid, which implies that $ AB$ is the perpendicular bisector of $ CX$. Since $ AOBX$ is cyclic we infer that \[ 180^{\cdot} = \angle{AOB} + \angle{AXB} = 2\angle{C} + \angle{C}, \] hence $ \angle{C} = 60^{\cdot}$.

Sorry for bumping this post up like 10 years, but can someone explain how $ COKX$ is a cyclic trapezoid in bravado's solution.