3. [15] Let $A B C$ be a triangle with circumcenter $O$ such that $A C=7$. Suppose that the circumcircle of $A O C$ is tangent to $B C$ at $C$ and intersects the line $A B$ at $A$ and $F$. Let $F O$ intersect $B C$ at $E$. Compute $B E$.
Answer: $E B=\frac{7}{2} \quad O$ is the circumcenter of $\triangle A B C \Longrightarrow A O=C O \Longrightarrow \angle O C A=\angle O A C$. Because $A C$ is an inscribed arc of circumcircle $\triangle A O C, \angle O C A=\angle O F A$. Furthermore $B C$ is tangent to circumcircle $\triangle A O C$, so $\angle O A C=\angle O C B$. However, again using the fact that $O$ is the circumcenter of $\triangle A B C, \angle O C B=\angle O B C$.
We now have that $C O$ bisects $\angle A C B$, so it follows that triangle $C A=C B$. Also, by AA similarity we have $E O B \sim E B F$. Thus, $E B^{2}=E O \cdot E F=E C^{2}$ by the similarity and power of a point, so $E B=B C / 2=A C / 2=7 / 2$.