10. On the side $AC$ of $\triangle ABC$, take a point $D$ such that $\angle BDC = \angle ABC$. If $BC = 1$, what is the minimum possible distance between the circumcenter of $\triangle ABC$ and the circumcenter of $\triangle ABD$?
10. $\frac{1}{2}$.

Considering only the case where $\triangle A B C$ is an acute triangle (as shown in Figure 5), other cases can be discussed similarly.
Figure 5
Let the circumcenters of $\triangle A B C$ and $\triangle A B D$ be $O_{1}$ and $O_{2}$, respectively.
$$
\begin{array}{l}
\text { By } \angle B D C=\angle A B C \\
\Rightarrow \triangle A B C \backsim \triangle B D C \\
\Rightarrow A C \cdot D C=B C^{2}=1 .
\end{array}
$$

Also, $\angle B O_{1} C=2 \angle B A C=\angle B O_{2} D$, then
$$
\begin{array}{l}
\triangle D B O_{2} \backsim \triangle C B O_{1} \\
\Rightarrow \angle O_{2} B D=\angle O_{1} B C .
\end{array}
$$

Since $O_{1} O_{2}$ is the perpendicular bisector of segment $A B$
$$
\begin{aligned}
\Rightarrow & \angle O_{1} O_{2} B=\frac{1}{2} \angle A O_{2} B \\
& =\frac{1}{2}\left(360^{\circ}-2 \angle A D B\right) \\
& =180^{\circ}-\angle A D B=\angle B D C, \\
& \angle O_{2} B O_{1}=\angle O_{2} B D+\angle D B O_{1} \\
& =\angle O_{1} B C+\angle D B O_{1}=\angle D B C .
\end{aligned}
$$

Thus, $\triangle O_{2} O_{1} B \backsim \triangle D B C$
$$
\begin{aligned}
\Rightarrow & O_{2} O_{1}=\frac{D C \cdot B O_{1}}{B C} \\
& =D C \cdot B O_{1}=\frac{D C}{2} \cdot 2 B O_{1} \\
& =\frac{D C}{2}\left(A O_{1}+O_{1} C\right) \\
& \geqslant \frac{D C \cdot A C}{2}=\frac{1}{2},
\end{aligned}
$$

The equality holds when point $O_{1}$ lies on segment $A C$, i.e., when $\triangle A B C$ is a right triangle.