We are working with directed angles modulo $180^{\circ}$.
In this sketch, $T$ corresponds to the point $P$.
!

Step 1: First, we observe that $C A$ and $C B$ are the perpendicular bisectors of the segments $\overline{P B_{1}}$ and $\overline{P A_{1}}$, so $C$ is the circumcenter of triangle $P A_{1} B_{1}$, and it follows that

$$\measuredangle A_{1} C B = \frac{1}{2} \measuredangle A_{1} C P = \measuredangle A_{1} B_{1} P = \measuredangle A_{1} B_{1} B_{2} = \measuredangle A_{1} C_{2} B_{2}.$$

Step 2: Analogously to Step 1, it follows that $\measuredangle A_{1} B C = \measuredangle A_{1} B_{2} C_{2}$, so triangles $A_{1} B C$ and $A_{1} B_{2} C_{2}$ are similar.

Step 3: Using the similarity just proven, we have

$$\measuredangle C A_{1} B = \measuredangle C_{2} A_{1} B_{2} \quad \text{and} \quad \frac{\left|\overline{A_{1} C}\right|}{\left|\overline{A_{1} B}\right|} = \frac{\left|\overline{A_{1} C_{2}}\right|}{\left|\overline{A_{1} B_{2}}\right|}$$

and thus also

$$\measuredangle B_{2} A_{1} B = \measuredangle C_{2} A_{1} C \quad \text{and} \quad \frac{\left|\overline{A_{1} B_{2}}\right|}{\left|\overline{A_{1} B}\right|} = \frac{\left|\overline{A_{1} C_{2}}\right|}{\left|\overline{A_{1} C}\right|}$$

so triangles $A_{1} B_{2} B$ and $A_{1} C_{2} C$ are also similar.
Step 4: Let $K$ be the intersection of $C_{2} C$ with $\Omega$. Then $180^{\circ} - \measuredangle A_{1} B_{2} K = \measuredangle A_{1} C_{2} K = \measuredangle A_{1} C_{2} C = \measuredangle A_{1} B_{2} B$, so $K$ lies on the line $B B_{2}$. Similarly, one can show that $K$ also lies on the line $A A_{2}$, which completes the proof.