Connect $A D$. From $\overparen{A E D}=\overparen{A B}$, we know
$$\angle A B D=\angle A D B,$$

and the extension of $B D$ intersects the extension of $A E$.
When $A$, $E$, and $Q$ are collinear, the extension of $B D$ intersects the extension of $A E$ at point $Q$. Then
$\angle D E Q=\angle A B D=\angle A D B$.
Thus, $\angle A E D=\angle A D Q$
$\Rightarrow \triangle A E D \backsim \triangle A D Q \Rightarrow \frac{A E}{A D}=\frac{A D}{A Q}$.
Also, $\angle A D P=\angle A C D$
$\Rightarrow \triangle A D P \backsim \triangle A C D \Rightarrow \frac{A P}{A D}=\frac{A D}{A C}$.
Therefore, $A E \cdot A Q=A D^{2}=A P \cdot A C$.
Hence, $C$, $P$, $E$, and $Q$ are concyclic.
When $C$, $P$, $E$, and $Q$ are concyclic, let the circumcircle of $\triangle C P E$ be $\odot O$, then $B D$ intersects $\odot O$ at points $P$ and $Q$.

If $A$, $E$, and $Q$ are not collinear, then the intersection point $Q^{\prime}$ of $B D$ and the extension of $A E$ is different from $Q$.

From the above proof, $C$, $P$, $E$, and $Q^{\prime}$ are concyclic, i.e., $Q^{\prime}$ is also on $\odot O$. Thus, line $B D$ intersects $\odot O$ at three points $P$, $Q$, and $Q^{\prime}$, which is impossible.
Therefore, $A$, $E$, and $Q$ are collinear.