My solution: Since $ OO_2 $ is the perpendicular bisector of $ AC_1 $ , so from $ OO_2 \perp AO_1 $ we get $ A, O_1, C_1 $ are collinear . Since $ \angle O_1CO=\angle OAO_1=\angle O_1C_1O $ so we get $ O, O_1, C, C_1 $ are concyclic . ... $ (1) $ Since $ \angle AB_1C=\angle ABC-\angle B_1CB=\angle ABC-\angle CAB $ , so $ \angle CAC_1=90^{\circ}-\angle AB_1C=90^{\circ}+\angle CAB-\angle ABC $ , hence we get $ \angle C_1AB=\angle CAB-\angle CAC_1=\angle ABC-90^{\circ} $ , so $ \angle AB_2C_1=180^{\circ}-\angle C_1CA-\angle C_1AB=180^{\circ}+\angle CAB-\angle ABC $ . Since $ \angle O_2AC_1=\angle AB_2C_1-90^{\circ}=90^{\circ}+\angle CAB-\angle ABC=\angle CAC_1 $ , so we get $ O_2 \in AC $ and $ \angle O_1OO_2=\angle CAO_1=\angle O_1CO_2 $ . ie. $ O, O_1, O_2, C $ are concyclic ... $ (2) $ From $ (1) $ and $ (2) $ we get $ O, O_1, O_2, C, C_1 $ are concyclic. Q.E.D

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