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Circle $\Omega$ is tangent to circles $\omega_1$ and $\omega_2$ at points $A_1$ and $A_2$, respectively. Let $B$ be a point on $\Omega$, let $K_1$ and $K_2$ be the other intersection points of lines $A_1B$ and $A_2B$ with $\omega_1$ and $\omega_2$, respectively. Prove that if $K_1K_2$ is tangent to $\omega_1$, then it is also tangent to $\omega_2$.

Please help with this problem, I tried using directed angles but made no progress at all

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Suppose $\omega_1$ touches $K_1K_2$.

Apply inversion with center at $B$ and $r=\sqrt{BK_1\cdot BA_1}$.

Then this inversion takes $K_1$ to $A_1$ and vice versa and thus it takes $\omega_1$ to it self. Since line $K_1K_2$ is tangent to $\omega _1$ it image, which is some circle through $B$ and $A_1$ also touches $\omega _1$. But then this is exactly $\Omega$. Now we know that $\Omega $ and $\omega _2$ are tangent to each other, therefore there images are also tangent to each other. Say image of $\omega _2$ is $\omega _2'$ and we know that image of $\Omega$ is $K_1K_2$. Thus $K_1K_2$ touches $\omega _2'$.

Now since $A_2$ is on $\Omega $ and on $A_2K_2$ it image is on image of this two, thus on $K_1K_2$ and $A_2K_2$, so the image of $A_2$ is $K_2$ (and vice versa) which means that $\omega _2' = \omega _2$. Q.E.D.

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Suppose $\omega_1$ touches $K_1K_2$.

Homothety at $A_1$ which takes $\omega _1$ to $\Omega$ takes $K_1$ to $B$ and $K_1K_2$ to (parallel) tangent $t$ to $\Omega$ at $B$. Homothety at $A_2$ which takes $\Omega $ to $\omega _2$ takes $B$ to $K_2$ and $t$ to (parallel) tangent $\ell$ to $\omega _2$ at $K_2$. Since $\ell|| K_1K_2$ we have $\ell = K_1K_2$. Q.E.D.

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