China-TST-B

Proof 1. Consider the antipodal point $P$ of $A$ on the circle $\odot O$.
We notice that in the right trapezoid $A_1A_2O_1O_2$, we have $O_1A_1 = O_1B_1$ and $OB_1 = OA$.
Therefore,
$$\begin{align*} \angle A_1BA &= \pi - \angle O_1B_1A_1 - \angle OB_1A \\ &= \pi - \frac{1}{2}(\pi - \angle A_1O_1O) - \frac{1}{2}(\pi - \angle AOO_1) \\ &= \frac{1}{2}(\angle A_1O_1O + \angle AOO_1) = \frac{\pi}{2}, \end{align*}$$
which implies that the line $A_1B_1$ passes through point $P$. Similarly, $A_2B_2$ also passes through point $P$.
We have,
$$\begin{align*} \angle D_1B_2B_1 &= \angle PB_2B_1 = \angle PAB_1 \quad (P, B_2, A, B_1 \text{ concyclic}) \\ &= \angle B_1A_1A = \angle B_1CA_1 \quad (\ell \text{ is tangent to } \odot O_1) \\ &= \angle B_1CD_1. \end{align*}$$
Hence, points $C, D_1, B_1, B_2$ are concyclic. Similarly, $D_2$ also lies on the circumcircle of $\triangle CB_1B_2$. Therefore,
$$\angle D_1D_2B_1 = \angle D_1B_2B_1 = \angle B_1A_1A.$$
This implies that $D_1D_2$ is parallel to $\ell$. Proof completed. $\Box$

Proof 2. We establish a coordinate system with $\ell$ as the $x$-axis and $A$ as the origin. Without loss of generality, let $O(0, 1)$. For $i = 1, 2$, let $r_i$ be the radius of $\odot O_i$, which corresponds to the $y$-coordinate of point $O_i$. From the equation
$$x_{O_i}^2 + (r_i - 1)^2 = (r_i + 1)^2,$$
we obtain $O_1(-2\sqrt{r_1}, r_1)$ and $O_2(2\sqrt{r_2}, r_2)$ (thus $A_1(-2\sqrt{r_1}, 0)$ and $A_2(2\sqrt{r_2}, 0)$). Furthermore, due to the tangent property of $\odot O_1$ and $\odot O_2$, we have
$$[2(\sqrt{r_1} + \sqrt{r_2})]^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2,$$
which simplifies to
$$(*) \qquad \sqrt{r_1} + \sqrt{r_2} = \sqrt{r_1} \sqrt{r_2}.$$
On the other hand, we have
$$\overrightarrow{AB_1} = \frac{1 \cdot \overrightarrow{AO_1} + r_1 \cdot \overrightarrow{AO}}{r_1 + 1},$$
which yields $B_1(\frac{-2\sqrt{r_1}}{r_1+1}, \frac{2r_1}{r_1+1})$. Similarly, we have $B_2(\frac{2\sqrt{r_2}}{r_2+1}, \frac{2r_2}{r_2+1})$ and $C(\frac{2(r_1\sqrt{r_2}-r_2\sqrt{r_1})}{r_1+r_2}, \frac{2r_1r_2}{r_1+r_2})$.

$$y = \frac{\frac{2r_1}{r_1+1}}{\frac{-2\sqrt{r_1}}{r_1+1} + 2\sqrt{r_1}}(x + 2\sqrt{r_1}) = \frac{1}{\sqrt{r_1}}x + 2,$$
and the equation of line $CA_2$ is given by
$$y = \frac{\frac{2r_1r_2}{r_1+r_2}}{\frac{2(r_1\sqrt{r_2}-r_2\sqrt{r_1})}{r_1+r_2} - 2\sqrt{r_2}} (x - 2\sqrt{r_2}),$$
which can be combined to obtain
$$y = \frac{2r_1r_2}{2\sqrt{r_1r_2}(\sqrt{r_1} - \sqrt{r_2}) - 2\sqrt{r_2}(r_1 + r_2)}[\sqrt{r_1}(y - 2) - 2\sqrt{r_2}].$$
Rearranging the terms, we obtain a linear equation in terms of $y$, with the constant term
$$4r_1r_2(\sqrt{r_1} + \sqrt{r_2}),$$
which is a symmetric expression in terms of $r_1$ and $r_2$. The coefficient of $y$ in this case is (using (*) to convert the expression into a homogeneous form)
$$\begin{aligned} & 2\{r_1r_2\sqrt{r_1} - \sqrt{r_1r_2}(\sqrt{r_1} - \sqrt{r_2}) + \sqrt{r_2}(r_1 + r_2)\} \\ &= 2\{(\sqrt{r_1} + \sqrt{r_2})^2\sqrt{r_1} - \sqrt{r_1r_2}(\sqrt{r_1} - \sqrt{r_2}) + \sqrt{r_2}(r_1 + r_2)\} \\ &= 2\{r_1\sqrt{r_1} + 2r_1\sqrt{r_2} + r_2\sqrt{r_1} - r_1\sqrt{r_2} + r_2\sqrt{r_1} + r_1\sqrt{r_2} + r_2\sqrt{r_2}\} \\ &= 2\{r_1\sqrt{r_1} + 2r_1\sqrt{r_2} + 2r_2\sqrt{r_1} + r_2\sqrt{r_2}\}, \end{aligned}$$
which is also a symmetric expression in terms of $r_1$ and $r_2$.
Therefore, we know that the $y$-coordinates of points $D_1$ and $D_2$ have the same expression, which proves the proposition. $\square$

Proof 3. (In fact, it is not necessary for circle $O$ to be tangent to line $\ell$.)
It is easy to see that $O, B_1, O_1$ are collinear, $O, B_2, O_2$ are collinear, $O_1, C, O_2$ are collinear, $OB_1 = OB_2$, $O_1A_1 = O_1B_1$, $O_2A_2 = O_2B_2$, and $O_1A_1, O_2A_2$ are both perpendicular to $\ell$.
Now, we have:
$$\begin{aligned} \angle CD_1B_2 &= \angle CA_1A_2 + \angle A_1A_2B_2 \\ &= \angle CA_1B_1 + \angle B_1A_1A_2 + \angle A_1A_2B_2 \\ &= \frac{1}{2}\angle CO_1B_1 + \frac{1}{2}\angle A_1O_1B_1 + \frac{1}{2}\angle A_2O_2B_2 \\ &= (90^\circ - \angle O_1B_1C) + \frac{1}{2}\angle O_1OO_2 \\ &= (\angle OB_1C - 90^\circ) + 90^\circ - \angle OB_1B_2 \\ &= \angle CB_1B_2. \end{aligned}$$
Therefore, $C, D_1, B_1, B_2$ are concyclic. Similarly, $C, D_2, B_2, B_1$ are concyclic. Hence, $C, D_1, B_1, B_2, D_2$ are concyclic. Furthermore, $\angle CD_1D_2 = \angle CB_1D_2 = 180^\circ - \angle A_1B_1C = \angle CA_1A_2$. Thus, we have $D_1D_2//\ell$. $\square$