The diagram shows two circles, $C_1$ and $C_2$, intersecting at $A$ and at $B$. $F$ is a point on $AB$ produced such that $FD$ and $FE$ are tangents to $C_1$ at $D$ and $C_2$ at $E$ respectively. $DBE$ is a straight line.

1. Identify the common angle in both triangles

$\angle DFB = \angle AFD$ (Common angle)

2. Use the alternate segment theorem to find a second pair of equal angles

Since $FD$ is a tangent to circle $C_1$ at $D$ and $DB$ is a chord, the angle between the tangent and the chord is equal to the angle in the alternate segment.
$\angle FDB = \angle DAB$ (Alternate segment theorem)
Since $F, B, A$ are collinear, $\angle DAB$ is the same as $\angle FAD$.
Therefore, $\angle FDB = \angle FAD$.

3. Conclude that the triangles are similar

Since two corresponding angles are equal, $\triangle FDB$ is similar to $\triangle FAD$ (AA similarity).

1. Identify the other pair of similar triangles

In $\triangle FEB$ and $\triangle FAE$:
$\angle EFB = \angle AFE$ (Common angle)
$\angle FEB = \angle FAE$ (Alternate segment theorem for circle $C_2$)
Therefore, $\triangle FEB$ is similar to $\triangle FAE$.

2. Establish the relationship between the sides of the newly identified similar triangles

Since $\triangle FEB$ and $\triangle FAE$ are similar, the ratio of their corresponding sides is equal:
$\frac{FE}{FA} = \frac{FB}{FE}$
$FE^2 = FA \times FB$

3. Use the result from part (i) to show $FD = FE$

From part (i), $\triangle FDB$ and $\triangle FAD$ are similar, so:
$\frac{FD}{FA} = \frac{FB}{FD}$
$FD^2 = FA \times FB$
Since $FD^2 = FA \times FB$ and $FE^2 = FA \times FB$, we have:
$FD^2 = FE^2$
Since $FD$ and $FE$ are lengths, $FD = FE$.

1. Deduce the angles at the intersection $B$.

Since the line $ABF$ is perpendicular to $DE$ and they intersect at $B$, we have $\angle FBD = 90^\circ$ and $\angle FBE = 90^\circ$.

2. Use the similarity from part (i) to find $\angle FDA$.

From (i), $\triangle FDB$ is similar to $\triangle FAD$.
Therefore, corresponding angles are equal: $\angle FDA = \angle FBD = 90^\circ$.

3. Use the similarity from part (ii) to find $\angle FEA$.

From (ii), $\triangle FEB$ is similar to $\triangle FAE$.
Therefore, corresponding angles are equal: $\angle FEA = \angle FBE = 90^\circ$.

4. Conclude using circle properties.

Since $\angle FDA = 90^\circ$ and $\angle FEA = 90^\circ$, the points $D$ and $E$ subtend a right angle from the line segment $AF$. By the converse of the angle in a semicircle theorem, a circle with diameter $AF$ must pass through $D$ and $E$.