Let $AD \cap \Omega = P$. Now, note that $AP = BB_1, AA_1 = CA_2$ and $BB_1 = CB_2$. Now, power of point implies that $AP.AD = AC.AA_1 \Rightarrow BB_1.BC = CA_2.AC \Rightarrow CA_2.AC=CB_2.BC$. Therefore, $A,B,A_2,B_2$ are cyclic.

Somebody post the next problem.