Let $\angle B A D=\angle A C D=\alpha, \angle B A F=\angle C A E=\beta$. Since point A lies on the circle with diameter $B C$, then $\angle B A C=90^{\circ}$. Therefore,

$$
\begin{gathered}
\angle A D C=180^{\circ}-\angle A C D-\angle C A D=180^{\circ}-\alpha-\left(90^{\circ}-\alpha\right)=90^{\circ}, \\
\angle F A E=\angle F A B+\angle B A E=\beta+\left(90^{\circ}-\beta\right)=90^{\circ} .
\end{gathered}
$$

Then $A D$ is the common altitude of the right triangles $B A C$ and $F A E$. By the theorem of the altitude of a right triangle drawn from the vertex of the right angle,

$$
F D \cdot D E=AD^2=B D \cdot D C \text{. }
$$

Let $E C=x$, we get the equation

$$
(4+2) 3=2(3+x)
$$

from which we find that $x=6$. Therefore,

$$
B C=B E+E C=5+6=11 .
$$

## Answer

11.
8,9

On a circle, a point $A$ is taken, on the diameter $B C$ - points $D$ and $E$, and on its extension beyond point $B$ - point $F$. Find $B C$, if $\angle B A D=\angle A C D, \angle B A F=\angle C A E, B D=2, B E=5$ and $B F=4$.