For any point $X$ on line $A_{i+2}B_i$, we have:
$$\frac{d_{X-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = \frac{d_{X-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}}.$$
For any point $X$ on line $A_iB_i$, we have:
$$\frac{d_{X-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{X-A_iA_{i+1}}}{A_iA_{i+1}} = 0.$$
Therefore, for point $B_i$:
$$\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}, \qquad (6)$$
and
$$\frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} = 0. \qquad (7)$$
Combining these gives:
$$\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} + \frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.$$
Similarly for $B_{i+1}$:
$$\frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} = \frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}},$$
and
$$\frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = 0.$$
Thus:
$$\frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.$$
Since these relations are linear in the coordinates, point $C_i$ must also satisfy:
$$\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} + \frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \qquad (8)$$

Noting that $C_i$ lies on $A_iA_{i+1}$ (so $d_{C_i-A_iA_{i+1}} = 0$), we can rewrite this as:
$$\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} + \frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} = \frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \quad (9)$$
This equation is completely symmetric for $C_1, C_2, C_3, C_4$. If it held identically for all points, then substituting $A_1$ would yield:
$$\frac{A_1 A_2 \cdot \sin A_2}{A_2 A_3} = \frac{A_1 A_4 \cdot \sin A_4}{A_3 A_4},$$
and substituting $A_3$ would give:
$$\frac{A_2 A_3 \cdot \sin A_2}{A_1 A_2} = \frac{A_3 A_4 \cdot \sin A_4}{A_1 A_4}.$$
This would imply $\sin A_2 = \sin A_4$, and similarly $\sin A_1 = \sin A_3$, meaning $A_1A_2A_3A_4$ would either be a parallelogram or cyclic - contradicting the given conditions.
Therefore, the equation is not identically satisfied. Since it is linear in coordinates, the set of points satisfying it must form a straight line. Hence $C_1, C_2, C_3, C_4$ are collinear. $\square$

Second Proof: Define the cross ratio of four lines $PA, PB, PC, PD$ through point $P$ as:
$$P(A, B; C, D) := \frac{\sin \angle APC \cdot \sin \angle BPD}{\sin \angle APD \cdot \sin \angle BPC}.$$
First, we prove a lemma.
Lemma: Let $A, B, C, D$ be four points in the plane with no three collinear, and let $M$ be another point. Then a point $P$ lies on the conic through $A, B, C, D, M$ if and only if $P(A, B; C, D) = M(A, B; C, D)$.
Proof of Lemma: Let $a, b, c, d, m, p$ be the complex coordinates of points $A, B, C, D, M, P$ respectively. Then:
$$\begin{aligned} \frac{PA \cdot PB}{PC \cdot PD} &= \frac{\sin \angle APB \cdot \sin \angle BPD}{\sin \angle ABP \cdot \sin \angle BPD} \\ &= \frac{\operatorname{Im}\left(\frac{c-p}{a-p} / \left|\frac{c-p}{a-p}\right|\right) \cdot \operatorname{Im}\left(\frac{d-p}{b-p} / \left|\frac{d-p}{b-p}\right|\right)}{\operatorname{Im}\left(\frac{c-p}{b-p} / \left|\frac{c-p}{b-p}\right|\right) \cdot \operatorname{Im}\left(\frac{d-p}{a-p} / \left|\frac{d-p}{a-p}\right|\right)} \\ &= \frac{[(c-p)(\bar{a}-\bar{p}) - (\bar{c}-\bar{p})(a-p)][(d-p)(\bar{b}-\bar{p}) - (\bar{d}-\bar{p})(b-p)]}{[(c-p)(\bar{b}-\bar{p}) - (\bar{c}-\bar{p})(b-p)][(d-p)(\bar{a}-\bar{p}) - (\bar{d}-\bar{p})(a-p)]} \end{aligned}$$
Since each bracket is linear in $p$ and $\bar{p}$, the condition "cross ratio equals a constant" defines a quadratic equation in $p$ and $\bar{p}$, which corresponds to a conic section. (Note: This cannot be identically constant, as that would imply $A, B, C$ are collinear, a contradiction.) $\square$

Returning to the problem, since $A_3B_1$ and $A_3M_1$ are symmetric about the angle bisector of $\angle A_2A_3A_4$, $A_3B_1$ is the symmedian of $\triangle A_2A_3A_4$.
Let $S$ be the intersection of $B_1A_1$ and $A_2A_4$, and $T$ the intersection of $B_1A_3$ and $A_2A_4$. Then:
$$\begin{aligned} B_1(A_1, A_2; A_3, A_4) &= (S, A_2; T, A_4) = 1 - (S, T; A_2, A_4) \\ &= 1 - \frac{SA_2}{SA_4} \cdot \frac{TA_4}{TA_2} = 1 - \frac{A_1A_2^2 \cdot A_3A_4^2}{A_1A_4^2 \cdot A_3A_2^2}. \end{aligned}$$
Similarly, this holds for $B_2, B_3, B_4$. Thus all eight points $A_1, \dots, A_4, B_1, \dots, B_4$ lie on the same conic.

* Considering $A_1, A_2, A_3, B_1, B_2, B_3$, by Pascal's Theorem, $C_1, C_2$ and $X := A_3B_1 \cap A_1B_3$ are collinear.
* Similarly, considering $A_1, A_4, A_3, B_1, B_4, B_3, C_3, C_4$ and $X$ are collinear.
* Considering $A_2, A_3, A_4, B_2, B_3, B_4, C_2, C_3$ and $Y := A_4B_2 \cap A_2B_4$ are collinear.
* Considering $A_4, A_1, A_2, B_4, B_1, B_2, C_1, C_4$ and $Y$ are collinear.

Let $W$ be the intersection of $A_1B_1$ and $A_1B_2$. Applying Menelaus' Theorem to
triangle $\triangle WB_1B_2$ with transversal $A_1A_2C_1$ gives:
$$\frac{B_1C_1}{B_2C_1} = \frac{B_1A_1}{B_2A_2} \cdot \frac{A_1W}{A_2W} = \frac{B_1A_1}{B_2A_2} \cdot \frac{\sin \angle WA_2A_1}{\sin \angle WA_1A_2}.$$
By tangent properties, $\angle WA_2A_1 = \angle A_2A_3A_1$, so:
$$\frac{B_1C_1}{B_2C_1} = \frac{B_1A_1}{B_2A_2} \cdot \frac{\sin \angle A_2A_3A_1}{\sin \angle A_1A_4A_2}.$$
Taking the cyclic product (indices modulo 4):
$$\begin{align} \prod_{i=1}^{4} \frac{B_i C_i}{B_{i+1} C_i} &= \prod_{i=1}^{4} \frac{B_i A_i}{B_{i+1} A_{i+1}} \cdot \prod_{i=1}^{4} \frac{\sin \angle A_{i+1}A_{i+2}A_i}{\sin \angle A_iA_{i-1}A_{i+1}} \tag{10} \\ &= \prod_{i=1}^{4} \frac{\sin \angle A_{i+1}A_{i+2}A_i}{\sin \angle A_{i+1}A_iA_{i+2}} = \prod_{i=1}^{4} \frac{A_i A_{i+1}}{A_{i+1} A_{i+2}} = 1. \nonumber \end{align}$$

Thus:
$$\frac{B_1 C_1}{B_2 C_1} \cdot \frac{B_2 C_2}{B_3 C_2} = \frac{B_1 C_4}{B_4 C_4} \cdot \frac{B_3 C_4}{B_3 C_3}.$$
Let this ratio be $\alpha$.
On the extension of $B_3B_1$, take point $P$ such that $\frac{B_3P}{B_1P} = \frac{1}{\alpha}$. By the converse of Menelaus' Theorem applied to $\triangle B_1B_2B_3$ and points $P, C_1, C_2$, these are collinear. Similarly, $P, C_3, C_4$ are collinear.
Thus we have $P, C_1, C_2$ collinear and $P, C_3, C_4$ collinear, while $X, C_1, C_2$ and $X, C_3, C_4$ are also collinear. Since $P$ lies on $B_1B_3$ and $X = A_1B_3 \cap A_3B_1$, if $P = X$ then $A_1 = B_1$ or $A_3 = B_3$, making $A_1A_3$ a symmedian for both $\triangle A_1A_2A_4$ and $\triangle A_2A_3A_4$, which would imply $A_1A_2A_3A_4$ is harmonic - contradicting the non-cyclic condition.
Therefore $P \neq X$, and consequently $C_1, C_2, C_3, C_4$ all lie on the line $PX$. $\square$