, 2015

Let $\gamma_i$ cross the lines $A_iC_{i+1}$ and $A_iC_{i+2}$ again at $X_{i+2}$ and $Y_{i+1}$, respectively.

Since the homothety centred at $A_i$, transforming $\gamma_i$ into $\gamma$, sends $X_{i+2}$ to $C_{i+1}$ and $Y_{i+1}$ to $C_{i+2}$, the lines $X_{i+2}Y_{i+1}$ and $C_{i+1}C_{i+2}$ are parallel, so $A_iC_{i+1}/A_iC_{i+2} = X_{i+2}C_{i+1}/Y_{i+1}C_{i+2}$.

On the other hand, since $C_i$ has equal powers relative to $\gamma_{i+1}$ and $\gamma_{i+2}$, $C_iX_{i+1} \cdot C_iA_{i+2} = C_iB_i^2 = C_iY_{i+2} \cdot C_iA_{i+1}$, it follows that $C_iX_{i+1} = C_iB_i^2/C_iA_{i+2}$ and $C_iY_{i+2} = C_iB_i^2/C_iA_{i+1}$.

Hence $A_iC_{i+1}/A_iC_{i+2} = (B_{i+1}C_{i+1}/B_{i+2}C_{i+2})^2(A_iC_{i+2}/A_iC_{i+1})$, so $A_iC_{i+1}/A_iC_{i+2} = B_{i+1}C_{i+1}/B_{i+2}C_{i+2}$, and consequently $\prod_{i=0}^2 (A_iC_{i+1}/A_iC_{i+2}) = \prod_{i=0}^2 (B_{i+1}C_{i+1}/B_{i+2}C_{i+2}) = 1$. Since the sines of the angles $C_iA_iC_j$ are proportional to the lengths of the corresponding chords $A_iC_j$, the conclusion follows by Ceva's theorem in trigonometric form in the triangle $C_0C_1C_2$.

Alternative Solution:
Let the tangents to $\gamma$ at $A_i$ and $A_{i+1}$ meet at $D_{i+2}$ (fig 1), and notice that the latter has equal powers relative to $\gamma_i$ and $\gamma_{i+1}$ to deduce that it lies on their radical axis, $B_{i+2}C_{i+2}$. Consequently, the lines $C_iD_i$ are concurrent at the radical center of the $\gamma_i$, and the conclusion follows by the lemma below (see Figure 2).

Lemma. Let $P_0P_1P_2$ be a triangle, and let $T_i$ be the touchpoint of the side $P_{i+1}P_{i+2}$ and the incircle $\gamma$ of the triangle $P_0P_1P_2$. Let further $Z_i$ be a point on the arc $T_{i+1}T_{i+2}$ of $\gamma$ not containing $T_i$. Then the lines $T_iZ_i$ are concurrent if and only if the lines $P_iZ_i$ are concurrent.