The diagram shows a quadrilateral ABCD whose vertices lie on the circumference of the circle. The point E lies on the extended line CB such that AE is a tangent to the circle. CE and DA are parallel lines.

1. Relate angle BAE to an angle inside the circle using the tangent.

$\angle BAE = \angle ACB$ (Alternate Segment Theorem)

2. Relate angle ACB to angle CAD using parallel lines and conclude.

Since $CE \parallel DA$,
$\angle ACB = \angle CAD$ (Alternate angles)
Therefore, $\angle BAE = \angle CAD$.

1. State the first pair of equal angles from part (i)

From (i), $\angle BAE = \angle CAD$.

2. Identify a second pair of equal angles using parallel lines and cyclic quadrilateral properties

Since $CE \parallel DA$, $\angle BEA = \angle DAE$ (alternate angles).
Also, $\angle DAE = \angle DCA$ (angles in the alternate segment, since $AE$ is a tangent and $AD$ is a chord).
Therefore, $\angle BEA = \angle DCA$.

3. Conclude similarity

Since two pairs of corresponding angles are equal ($\angle BAE = \angle CAD$ and $\angle BEA = \angle DCA$), $\triangle BAE$ is similar to $\triangle DAC$ (AA similarity).

1. Use the given side equality to find equal angles in triangle ABE.

Given $AB = BE$, $\triangle ABE$ is an isosceles triangle.
Therefore, $\angle BAE = \angle BEA$.

2. Use the similar triangles from part (ii) to relate angles in triangle DAC.

From (ii), $\triangle BAE$ and $\triangle DAC$ are similar.
Corresponding angles are equal, so $\angle BEA = \angle ACD$.
Since $\angle BAE = \angle CAD$ (from part i) and $\angle BAE = \angle BEA$, it follows that $\angle CAD = \angle ACD$.

3. Use parallel lines to show the two parts of angle BCD are equal.

Given $CE \parallel DA$, alternate angles are equal, so $\angle BCA = \angle CAD$.
Substituting $\angle CAD = \angle ACD$ from the previous step gives $\angle BCA = \angle ACD$.
Hence, the line $AC$ bisects $\angle BCD$.