Does really $\mathcal{H}(\triangle A_1B_1C_1)=\triangle A_2B_2\mathcal{H}(C_1)$ ? And shouldn't the areas be proportional to squares of that?

My solution:

Maybe someone have a little more elegant solution? Mine sound a bit like brute force...

Circles $\omega_1$ and $\omega_2$ have no common point, have externals tangents $a$ and $b$, and internal tangent $c$. Lines $a, b$ and $c$ touches circle $\omega_1$ respectively on points $A_1$, $B_1$ and $C_1$, and circle $\omega_2$ respectively ay points $A_2$, $B_2$ and $C_2$. Prove that the ratio of the areas of the triangles $A_1B_1C_1$ and $A_2B_2C_2$ is the same as the ratio of radii of $\omega_1$ and $\omega_2$.