The diagram shows a quadrilateral $ABCD$ whose vertices lie on the circumference of a circle. The point $E$ lies on the extended line $CD$ such that $AE$ is a tangent to the circle at $A$. $CD$ and $AB$ are parallel lines.

1. Relate the interior opposite angle to the adjacent interior angle of the cyclic quadrilateral.

$\angle CDA + \angle CBA = 180^\circ$ (opposite angles of a cyclic quadrilateral)

2. Relate the exterior angle to the adjacent interior angle and conclude the proof.

$\angle CDA + \angle EDA = 180^\circ$ (angles on a straight line)
Therefore, $\angle CBA = 180^\circ - \angle CDA = \angle EDA$.
(Alternatively, state directly that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle).

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

From part (a), we have $\angle EDA = \angle CBA$.

2. Find a relationship for angle DAE using the alternate segment theorem

Since $AE$ is a tangent to the circle at $A$ and $AD$ is a chord, by the alternate segment theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment:
$\angle DAE = \angle ACD$

3. Relate angle ACD to angle BAC using parallel lines

Given that $CD$ is parallel to $AB$, the alternate angles are equal:
$\angle ACD = \angle BAC$
Therefore, $\angle DAE = \angle BAC$.

4. Conclude similarity using the AA criterion

In $\triangle DAE$ and $\triangle BAC$:

1. $\angle EDA = \angle CBA$ (proven)
2. $\angle DAE = \angle BAC$ (proven)

Since two pairs of corresponding angles are equal, $\triangle DAE$ and $\triangle BAC$ are similar (AA similarity).

1. Use the given equal sides to identify equal angles in triangle DAE.

In $\triangle DAE$, since $AD = DE$, it is an isosceles triangle.
Therefore, $\angle DAE = \angle DEA$ (base angles of isosceles triangle).

2. Use the similarity from part (b) to find equal angles in triangle BAC.

From part (b), $\triangle DAE$ is similar to $\triangle BAC$.
Corresponding angles of similar triangles are equal, so $\angle BAC = \angle DAE$ and $\angle BCA = \angle DEA$.
Since $\angle DAE = \angle DEA$, it follows that $\angle BAC = \angle BCA$.

3. Use parallel lines to relate the angles and conclude the proof.

Given that $AB$ is parallel to $CD$, we have $\angle BAC = \angle ACD$ (alternate angles).
Substituting this into the previous result gives $\angle BCA = \angle ACD$.
Therefore, the line $AC$ bisects $\angle BCD$.