In the diagram, $AB$ is the diameter of the circle with centre $O$. $DE$ and $BF$ are tangents to the circle at $C$ and $B$ respectively. $DCE$ and $BEF$ are straight lines.

1. Identify the right angle in triangle ABC

$\angle ACB = 90^\circ$ (angle in a semicircle, since $AB$ is the diameter).

2. Identify the right angle in triangle AFB

$\angle ABF = 90^\circ$ (tangent perpendicular to radius/diameter, since $BF$ is a tangent at $B$).

3. Conclude similarity using the shared angle and the right angles

In $\triangle ABC$ and $\triangle AFB$:

1. $\angle ACB = \angle ABF = 90^\circ$
2. $\angle CAB = \angle FAB$ (common angle)
   Therefore, $\triangle ABC$ and $\triangle AFB$ are similar (AA similarity).

1. Establish that $A, C, F$ are collinear.

From part (a), $\triangle ABC$ and $\triangle AFB$ are similar. Thus, $\angle BAC = \angle BAF$. Since $C$ and $F$ lie on the same side of $AB$, the points $A, C,$ and $F$ must be collinear.

2. Find $\angle ECF$ using the alternate segment theorem.

Since $DCE$ is a tangent to the circle at $C$ and $AC$ is a chord, by the alternate segment theorem, the angle between the tangent $CD$ and chord $AC$ is $\angle ACD = \angle ABC$.
Since $A, C, F$ is a straight line and $D, C, E$ is a straight line, $\angle ECF$ and $\angle ACD$ are vertically opposite angles.
Therefore, $\angle ECF = \angle ACD = \angle ABC$.

3. Show that $\triangle ECF$ is isosceles.

From the similarity of $\triangle ABC$ and $\triangle AFB$, we also have $\angle AFB = \angle ABC$.
Since $F$ is on the line $AF$ and $E$ is on $BF$, $\angle EFC$ is the same as $\angle AFB$.
Thus, $\angle EFC = \angle ABC$.
Comparing the angles, we get $\angle ECF = \angle EFC$.
Therefore, $\triangle ECF$ is an isosceles triangle, which means $EC = EF$.