A special configuration of Nine Circles Theorem and Eight Circle Theorem

Question

The result as follows from special configuration of merge Nine Circle Theorem and Eight Circle theorem but it is new:

Problem: Let three circle $(A)$, $(B)$, $(C)$ , let $A_c$ be arbitrary point in the circle $(C)$. Construct the circle $(O_1)$ through $A_c$ tangent to $(C)$ and $(B)$, denote $(O_1)$ tangent to $(B)$ at point $A_b$; Construct the circle $(O_2)$ through $A_b$ tangent to $(B)$ and $(A)$, denote $(O_2)$ tangent to $(A)$ at point $B_a$; Construct the circle $(O_3)$ through $B_a$ tangent to $(A)$ and $(C)$, denote $(O_3)$ tangent to $(C)$ at point $B_c$; Construct the circle $(O_4)$ through $B_c$ tangent to $(C)$ and $(B)$, denote $(O_4)$ tangent to $(B)$ at point $C_b$; Construct the circle $(O_5)$ through $C_b$ tangent to $(B)$ and $(A)$, $(O_5)$ tangent to $(A)$ at point $C_a$; Construct the circle $(O_6)$ through $C_a$ tangent to $(A)$ and $(C)$, then

1. By the Nine Circle Theorem $O_6$ also tangent to $(O_1)$ but new: the point of tangency is $A_c$.
2. Six points $A_c, A_b, B_a, B_c, C_b, C_a $ lie on a circle.

Remarks: This result 2. show that this configuration is special configuration of Eight Circle theorem, when two big circles coincide, please see Eight Circle theorem

3. Three lines $O_1O_4, O_2O_5, O_3O_6$ are concurrent;
4. Three lines $AO_1, BO_3, CO_5$ are concurrent
5. Three line $AO_2, BO_4, CO_6$ are concurrent
6. The points of concurrence in (3), (4), (5) are collinear.

Question: I am looking for a proof the result above