Question 412610
I was able to solve this problem entirely by angle chasing. However, drawing a diagram on this website is extremely difficult, so I'll try to be as verbose as possible (maybe another tutor will post a diagram). I recommend you draw a diagram on your paper, going step by step carefully.


Since ABCD are four points on a circle, then ABCD is cyclic. Suppose that:


{{{omega[1]}}} is the circle determined by A, B, A', B'.
{{{omega[2]}}} is the circle determined by B, C, B', C'.
{{{omega[3]}}} is the circle determined by C, D, C', D'.
{{{omega[4]}}} is the circle determined by D, A, D', A'.


Suppose that angle ADD' = {{{alpha}}}, angle D'DC = {{{beta}}}, angle CBB' = {{{gamma}}}, and angle ABB' = {{{theta}}}. Since ABCD is cyclic, then {{{alpha + beta + gamma + theta = 180}}}. Also, note that quadrilaterals ABB'A', BCC'B', CDD'C', and DAA'D' are all cyclic, as they are inscribed by circles {{{omega[1]}}}, ..., {{{omega[4]}}}, so:


angle AA'D' = {{{180 - alpha}}}
angle D'C'C = {{{180 - beta}}}
angle B'C'C = {{{180 - gamma}}}
angle B'A'A = {{{180 - theta}}}


Hence, we can find that angle B'C'D' = {{{360 - (180 - beta) - (180 - gamma) = beta + gamma}}} and angle B'A'D' = {{{360 - (180 - alpha) - (180 - theta) = alpha + theta}}}. Since {{{alpha + beta + gamma + theta = 180}}}, then angle B'C'D' + angle B'A'D' = 180. They are opposite angles, so by definition, A'B'C'D' is cyclic.