Let $\alpha = \angle{ADC}$, $\beta = \angle{BAA'}$, $\gamma = \angle{D'AD}$. $\angle{D'A'B'} = 2\pi - \angle{D'A'A} - \angle{AA'B} - \angle{BA'C} = 2\pi - \frac{1}{2}(\alpha + \beta - \gamma) - (\pi - 2\beta) - \frac{1}{2}(\pi - \alpha + \beta + \gamma) = \frac{1}{2}\pi + \beta$. $\angle{D'C'B'} = \angle{DC'C} - \angle{DC'D'} - \angle{CC'B'} = (\pi - 2\beta) - \frac{1}{2}(\pi - \alpha - \beta - \gamma) - \frac{1}{2}(\alpha + \gamma - \beta) = \frac{1}{2}\pi - \beta$. Since $\angle{D'A'B'} + \angle{D'C'B'} = \pi$, $(D'A'B'C')$ is cyclic.