, 2024

Let $\ell_1 \cap \ell_2 = P$. Since $\angle AXY = \angle YAC$ and $\angle CXY = \angle YCA$ we have $\angle AXC + \angle AYC = 180^\circ$. Therefore, by symmetry, we have $\angle AZC = \angle AYC = 180^\circ - \angle AXC$ hence $A, X, C, Z$ are concyclic.

Now we will prove that $PX$ is tangent to this circle. Let the second intersection point of the line $PX$ with the circle $\omega_1$ be $Q$. Since $P$ is the center of the homothety sending $\omega_1$ to $\omega_2$, it sends $A$ to $C$ and $Q$ to $P$. Therefore, we have $AQ \parallel XC$ and $\angle XCA = \angle QAP$ and by the tangency we have $\angle QAP = \angle QXA$ hence $\angle XCA = \angle PXA$ which implies the desired tangency.

Since $P$ lies on the symmetry axis of the two circles, we have $PX = PY = PZ = \sqrt{PA \cdot PC}$ and $PZ$ is also tangent to the circle $(AXCZ)$. Letting $XZ \cap AC = T$, we then have that $(A, C; P, T) = -1$. Then we have $(KA, KC; KP, KT) = -1$. Let the second intersection of $PK$ and $\omega_1$ be $S$. Then by carrying this harmonic bundle to the circle $\omega_1$ we see that $(A, B; S, KT \cap \omega_1) = -1$. On the other hand, we know that $(A, B; S, K) = -1$ hence $KT$ is the tangent to $\omega_1$ at $K$.

Similarly we can prove that $LT$ is the line passing through $L$ and tangent to $\omega_2$, hence the proof is completed.