在下图中，$BD$ 是圆 $ACB$ 的切线，且 $AB$ 是圆 $BCDE$ 的切线。$ACFE$ 是一条直线。

1. Relate the angle between the tangent $BD$ and chord $BC$ to the angle in the alternate segment.

Since $BD$ is a tangent to circle $ACB$ at $B$, by the alternate segment theorem,
$\angle DBC = \angle BAC$.

2. Relate $\angle DBC$ to another angle in the larger circle $BCDE$.

In circle $BCDE$, the angles subtended by chord $CD$ at the circumference are equal. Therefore, angles in the same segment give:
$\angle CBD = \angle CED$.

3. Conclude that $AB$ is parallel to $DE$ using alternate interior angles.

From the above, $\angle BAC = \angle CED$.
Since $A, C, F, E$ is a straight line, $\angle BAC$ is the same as $\angle BAE$, and $\angle CED$ is the same as $\angle AED$.
Thus, $\angle BAE = \angle AED$.
Since the alternate interior angles are equal, $AB \parallel DE$.

1. Show that triangle $BCF$ is similar to triangle $ABF$

In $\triangle ABF$ and $\triangle BCF$:
$\angle AFB = \angle BFC$ (Common angle)
$\angle BAF = \angle CBF$ (Alternate segment theorem, tangent $BD$ and chord $BC$)
$\therefore \triangle ABF \sim \triangle BCF$ (AA similarity)

2. Show that triangle $EDF$ is similar to triangle $ABF$

In $\triangle ABF$ and $\triangle EDF$:
$\angle AFB = \angle DFE$ (Vertically opposite angles)
$\angle ABF = \angle FDE$ (Alternate angles, $AB \parallel DE$)
$\therefore \triangle ABF \sim \triangle EDF$ (AA similarity)

1. Identify equal angles using the alternate segment theorem

Since $BD$ is a tangent to circle $ACB$ at $B$ and $BC$ is a chord, the angle between the tangent and the chord is equal to the angle in the alternate segment.
Therefore, $\angle CBF = \angle CAB$ (or $\angle BAF$).

2. Prove that triangle ABF is similar to triangle BCF

In $\triangle ABF$ and $\triangle BCF$:

1. $\angle BAF = \angle CBF$ (proven above)
2. $\angle AFB = \angle BFC$ (common angle, as $A, C, F$ are collinear)
   Thus, $\triangle ABF$ is similar to $\triangle BCF$.

3. Use the ratio of corresponding sides to show the required result

Since $\triangle ABF \sim \triangle BCF$, the ratio of their corresponding sides is equal:
$\frac{AF}{BF} = \frac{BF}{CF}$
Cross-multiplying gives:
$AF \times CF = BF^2$