EGMO Problem 3.20 (BAMO 2013/3)

I believe you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.

So $$AC' = AO = CO = CA'$$

and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.