In the diagram, $BC$ is a diameter of the circle. $ABC$ is a straight line and $AG$ is a tangent to the circle at point $F$. The line $DF$ intersects $BC$ at point $E$.

1. Identify the common angle between the two triangles.

$\angle FAB = \angle FAC$ (Common angle)

2. Use circle properties to find a second pair of equal angles.

$\angle AFB = \angle FCA$ (Alternate segment theorem)

3. Conclude the similarity of the triangles.

Since two corresponding angles are equal, $\triangle ABF$ and $\triangle AFC$ are similar (AA similarity).

1. Use the similarity of triangles ABF and AFC to set up a ratio of corresponding sides.

Since $\triangle ABF$ is similar to $\triangle AFC$ (from part (a)), the ratio of their corresponding sides is equal:
$\frac{AB}{AF} = \frac{AF}{AC}$

2. Rearrange the ratio to obtain the required equation.

Cross-multiplying gives:
$AF \times AF = AB \times AC$
$AF^2 = AB \times AC$