10. As shown in Figure 4, given that the radii of $\odot O_{1}$ and $\odot O_{2}$ are $r_{1}$ and $r_{2}$ respectively, $\odot O_{2}$ passes through point $O_{1}$, and the two circles intersect at points $A$ and $B$. Point $C$ is on $\odot O_{2}$, and line segment $A C$ intersects $\odot O_{1}$ at point $D$. Then, line segments $B C$, $B D$, $A O_{1}$, $A O_{2}$, and $O_{1} O_{2}$ are connected. There are the following four conclusions:
(1) $\angle B D C = \angle A O_{1} O_{2}$; (2) $\frac{B D}{B C} = \frac{r_{1}}{r_{2}}$;
(3) $A D = D C$; (4) $B C = D C$.
Among these, the correct conclusions are
10. (1) (2) (4).

As shown in Figure 8, extend $C B$ to intersect $\odot O_{1}$ at point $E$, and connect $A E$.

Since $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, we have
$$
\angle C = \angle A O_{2} O_{1}.
$$

Similarly, $\angle E = \angle A O_{1} O_{2}$.
Since $\angle B D C$ is the exterior angle of the cyclic quadrilateral $A E B D$ inscribed in $\odot O_{1}$, we have $\angle B D C = \angle E = \angle A O_{1} O_{2}$.
Therefore, conclusion (1) is correct.
Thus, $\triangle B C D \sim \triangle A O_{2} O_{1}$, which gives
$$
\frac{B D}{B C} = \frac{A O_{1}}{A O_{2}} = \frac{r_{1}}{r_{2}}.
$$

Therefore, conclusion (2) is correct.
Since $\odot O_{2}$ passes through point $O_{1}$, we have $\angle A O_{1} O_{2} = \angle O_{1} A O_{2}$. From $\triangle B C D \sim \triangle A O_{2} O_{1}$, we know $\angle D B C = \angle O_{1} A O_{2}$. Therefore, $\angle D B C = \angle B D C$, which implies $B C = D C$.
Therefore, conclusion (4) is correct.
When point $C$ moves along the circumference of $\odot O_{2}$ closer to point $B$, point $D$ also moves along the circumference of $\odot O_{1}$ closer to point $B$. At this time, $D C$ decreases, and $A D$ increases. Therefore, conclusion (3) is incorrect.