Example 1 Given that $AB$ is a chord of $\odot O$ with radius 1, and $AB=a<1$. Using $AB$ as one side, construct a regular $\triangle ABC$ inside $\odot O$, and let $D$ be a point on $\odot O$ different from point $A$, such that $DB=AB=a$, and the extension of $DC$ intersects $\odot O$ at point $E$. Then the length of $AE$ is ( ).
(A) $\frac{\sqrt{5}}{2} a$
(B) 1
(C) $\frac{\sqrt{3}}{2}$
(D) $a$
Solve As shown in Figure 2, connect $A D, O E, O A$.
Notice that
$$
D B=A B=B C \text {. }
$$

By Judgment 1, $B$ is the circumcenter of $\triangle A C D$.

Then, by Property 2(1), $\angle A D C=\frac{1}{2} \angle A B C$.
Thus, $O$ is the circumcenter of $\triangle A D E$.
Therefore, $\angle A D E=\frac{1}{2} \angle A O E$.
Hence, $\angle A O E=\angle A B C=60^{\circ}$.
Also, since $O E=O A$, $\triangle O A E$ is an equilateral triangle.
So, $A E=O A=1$.
Therefore, the answer is B.