Two circles with two common outer tangents have same chords Let $\omega_1$ and $\omega_2$ be two circles, $r_1 < r_2$, they have two common outer tangents, let $A$ and $B$ - common points of first outer tangent and $\omega_1$ and $\omega_2$ respectively, $C$ and $D$ - common points of second outer tangent, $E$ and $F$ are common points of line $BC$ and  $\omega_1$ and $\omega_2$ respectively. Prove, that $EC = BF$.




Join centers $G$ and $H$.  Since $ABHG$ and $CDHG$ are congruent trapezoids, then tangents $AB$ and $CD$ are equal.
But $AB^2=BE\cdot BC$, and $CD^2=CF\cdot CB$ [Euclid III, 36].
Therefore,$$BE\cdot BC=CF\cdot CB$$and$$\frac{BE}{CB}=\frac{CF}{BC}$$making$$BE=CF$$Therefore$$CE=BF$$ 
