Points A, B, C and D lie on a circle. The tangent to the circle at A meets BD produced at E. AE is parallel to BC.

1. Identify the relationship between the angle formed by the tangent and chord, and the angle in the alternate segment.

Let the tangent at $A$ be line $AE$. The angle between the tangent $AE$ and the chord $AB$ is $\angle BAE$.
By the alternate segment theorem, $\angle BAE = \angle BCA$.

2. Use the parallel lines property to find an alternate angle.

Since $AE \parallel BC$, the alternate angles are equal.
Therefore, $\angle BAE = \angle ABC$.

3. Equate the angles and conclude the triangle is isosceles.

From the previous steps, $\angle BCA = \angle BAE$ and $\angle ABC = \angle BAE$.
Thus, $\angle BCA = \angle ABC$.
In $\triangle ABC$, since the base angles are equal, the sides opposite to them are equal.
Therefore, $AB = AC$.

1. Express $\angle BAC$ in terms of $\angle ABC$.

Let $\angle ABC = x$.
From (i), $AB = AC$, so $\triangle ABC$ is an isosceles triangle.
Therefore, $\angle ACB = \angle ABC = x$.
Sum of angles in $\triangle ABC$:
$\angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 2x$.

2. Relate $\angle BAC$ to $\angle BDC$ using circle properties.

Angles subtended by the same arc $BC$ at the circumference are equal.
Therefore, $\angle BDC = \angle BAC = 180^\circ - 2x$.

3. Find $\angle CDE$ and conclude the proof.

Since $B, D, E$ lie on a straight line, $\angle CDE$ and $\angle BDC$ are supplementary.
$\angle CDE = 180^\circ - \angle BDC$
$\angle CDE = 180^\circ - (180^\circ - 2x) = 2x$
Since $x = \angle ABC$, we have $\angle CDE = 2\angle ABC$.