Example 6 In trapezoid $A B C D$, $A D / / B C, B C=$ $B D=1, A B=A C, C D<1, \angle B A C+\angle B D C$ $=180^{\circ}$ Find the value of $C D$.
Solution: As shown in Figure 7, construct: point $D$ as the symmetric point of $\mathrm{K} B C$, and connect $A E$, $B E$, and $C E$. Let $A E^{1} \cdot \mathrm{j}$ $B C$ intersect at point $F$.

If $A I) B C$, then points $A$ and $E$ are equidistant from $B C$, so,
$$
A F = F E.
$$

Let $C I) = C E = x$, $A F' = F E = m$.
Since $\angle B A C + \angle B D C = 180^{\circ}$, we have $\angle B A C + \angle B E C = 180^{\circ}$.
Therefore, $A, B, E, C$ are concyclic.
By $A B = A C$, we get $\angle A B C = \angle A C B$.
Thus, $\angle 1 = \angle A C B = \angle A B C = \angle 2$.
Also, $\angle E B F = \angle E A C$, so,
$\triangle B F E \sim \triangle A C E$.
Therefore, $\frac{B E}{F E} = \frac{A E}{C E}$.
Thus, $2 m^{2} = A E \cdot F E = B E \cdot C E = x$.
By the angle bisector property,
$$
\frac{B F}{C F} = \frac{B E}{C E} = \frac{1}{x}.
$$

Since $B F + C F = 1$, we have,
$$
B F = \frac{1}{x+1}, \quad C F = \frac{x}{x+1}.
$$

From equation (2) and the intersecting chords theorem, we get
$$
m^{2} = A F \cdot F' E' = B F \cdot F' C = \frac{x}{(x+1)^{2}}.
$$

Substituting equation (3) into equation (1), we get
$$
\frac{2 x}{(x+1)^{2}} = x.
$$

Solving for $x$, we get $x = \sqrt{2} - 1$.
Therefore, $C D = \sqrt{2} - 1$.