To this end, let the parallel through $O$ to the tangent $A'_{k+1}A'_{k+2}$ meet the tangents $A'_k A'_{k+1}$ and $A'_k A'_{k+2}$ at $X_k$ and $Y_k$, respectively, and notice that:

(1) the angles $OX_k B_k$ and $OA_{k+2} B_k$ are equal, since the points $A_{k+2}$, $B_k$, $O$, $X_k$ all lie on the circle on diameter $OX_k$;

(2) the angles $OA_{k+2}B_k$ and $OA_{k+1}A_{k+2}$ are equal, since $OA_{k+1} = OA_{k+2}$; and

(3) the angles $OA_{k+1}A_{k+2}$ and $OY_k B_k$ are equal, since the points $A_{k+1}$, $B_k$, $O$, $Y_k$ all lie on the circle on diameter $OY_k$.

Consequently, the angles $OX_k B_k$ and $OY_k B_k$ are equal. Since the lines $X_k Y_k$ and $A'_{k+1} A'_{k+2}$ are parallel, and the latter is perpendicular to the line $OA_k = OB_k$, it follows that $B_k$ is the midpoint of the segment $X_k Y_k$, whence the conclusion.

Alternative Solution:

We will show that $\prod_{k=0}^2 (C_k B_{k+1} / C_k B_{k+2}) = 1$, so the conclusion follows by Menelaus' theorem. To this end, we will prove that
$$\frac{C_k B_{k+1}}{C_k B_{k+2}} = \frac{A_k A_{k+1}}{A_k A_{k+2}} \cdot \frac{A_k B_{k+1}}{A_k B_{k+2}}. \quad (*)$$
Multiplying the three yields the desired relation, since, by Ceva's theorem, $\prod_{k=0}^2 (A_k B_{k+1} / A_k B_{k+2}) = 1$.

We now turn to prove $(*)$. To avoid directed angles, assume, without any loss, that the triangle $A_0 A_1 A_2$ is acute-angled, fix an index $k$ and let $A_k A_{k+1} \leq A_k A_{k+2}$ (the case $A_k A_{k+2} \leq A_k A_{k+1}$ is dealt with similarly). Apply the law of sines in the triangles $A_k B_{k+1} C_k$ and $A_k B_{k+2} C_k$ to get
$$\frac{C_k B_{k+1}}{C_k B_{k+2}} = \frac{A_k B_{k+1}}{A_k B_{k+2}} \cdot \frac{\sin(\angle C_k A_k B_{k+1})}{\sin(\angle C_k A_k B_{k+2})}.$$
Since the angle $C_k A_k B_{k+1}$ is the supplement of the internal angle at $A_{k+1}$ of the triangle $A_0 A_1 A_2$, and the angle $C_k A_k B_{k+2}$ is equal to the internal angle at $A_{k+2}$ of the triangle $A_0 A_1 A_2$, the above ratio of sines equals $A_k A_{k+1} / A_k A_{k+2}$ and $(*)$ follows.