点 A、B、C 和 D 在一个圆上。在点 A 处的圆切线与 BD 的延长线相交于点 E。AE 平行于 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 $\angle CDE = 2\angle ABC$.