By $\angle (\boldsymbol{l}, n)$ we always mean the directed angle of the lines $l$ and $n$, taken modulo $180^\circ$.

Let $CC_2$ meet $\Omega$ again at $K$ (as usual, if $CC_2$ is tangent to $\Omega$, we set $K = C_2$). We show that the line $BB_2$ contains $K$; similarly, $AA_2$ will also pass through $K$. For this purpose, it suffices to prove that
$$\angle(C_2C, C_2A_1) = \angle(B_2B, B_2A_1). \qquad (1)$$
By the problem condition, $CB$ and $CA$ are the perpendicular bisectors of $TA_1$ and $TB_1$, respectively. Hence, $C$ is the circumcenter of the triangle $A_1TB_1$. Therefore,
$$\angle(CA_1, CB_1) = \angle(CB, CT) = \angle(B_1A_1, B_1T) = \angle(B_1A_1, B_1B_2).$$

In circle $\Omega$ we have $\angle(B_1A_1, B_1B_2) = \angle(C_2A_1, C_2B_2)$. Thus,
$$\angle(CA_1, CB) = \angle(B_1A_1, B_1B_2) = \angle(C_2A_1, C_2B_2). \qquad (2)$$
Similarly, we get
$$\angle(BA_1, BC) = \angle(C_1A_1, C_1C_2) = \angle(B_2A_1, B_2C_2). \qquad (3)$$
The two obtained relations yield that the triangles $A_1BC$ and $A_1B_2C_2$ are similar and equioriented, hence
$$\frac{A_1B_2}{A_1B} = \frac{A_1C_2}{A_1C} \quad \text{and} \quad \angle(A_1B, A_1C) = \angle(A_1B_2, A_1C_2).$$
The second equality may be rewritten as $\angle(A_1B, A_1B_2) = \angle(A_1C, A_1C_2)$, so the triangles $A_1BB_2$ and $A_1CC_2$ are also similar and equioriented. This establishes (1). $\Box$