Congruent Segments Associated to Tangent Lines.

Let $A$ and $B$ be the centers two circles $(A)$, $(B)$. From $A$, draw a tangent to $(B)$ at $C$. From $B$, draw two tangents to $(A)$ in $D$, $E$. From $C$, draw $CD$ such that intersect $(A)$ again in $F$. Similarly, from $C$, draw $CE$ such that intersect $(A)$ in $I$.

Prove $CF = CE$.

Proof.

Since $BD$ and $AC$ are tangent lines it follows that $\angle{BDA}=\angle{ACB}=\angle{AEB}=90^\circ$, hence $A$, $B$, $C$, $D$, $E$ are concyclic. Now, see that $\angle{DCE}=\angle{DBE}$ and $\angle{ACE}=\angle{ABE}$. $\angle{ABE}=\frac{\angle{DBE}}{2}=\frac{\angle{DCE}}{2}$, then, segment $AC$ is an angle bisector of $\angle{DCE}$. Focus on $\triangle{ACF}$. $\angle{CDB}=\angle{CEB}$. $\angle{FDA}=\angle{DFA}=180^\circ-(90^\circ+\angle{CDB})=90^\circ-\angle{CDB}$.$\angle{FAC}= 180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})=90^\circ+\angle{CDB}-\angle{DCA}$. Focus on $\triangle{ACE}$. $\angle{EAC}=180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})$. $\angle{EAC}=90^\circ+\angle{CDB}-\angle{DCA}$. Hence $\angle{FAC}=\angle{EAC}$, and $\triangle{ACF}$, $\triangle{ACE}$ are congruent. Thus, $CF=CE$.