Is this a similarity criterion?

Unfortunately, the given fact of the porpotionality of two sides and the circumradii of the triangles, does not guarantee similarity. The triangles may be similar, as well as they may not.

In figures 1 and 2 , we have triangles $\triangle ABC$ and $\triangle A_1 B_1 C_1$ with sides $AB=c$ , $AC=b$ , $A_1B_1=c_1$ , $A_1C_1=b_1$ , and circumradii $R$ , $R_1$ respectively.
Suppose $\frac{b}{{{b_1}}} = \frac{c}{{{c_1}}} = \frac{R}{{{R_1}}}.$

Then, by SSS (side-side-side) criterion, $\triangle OAB$ is similar to $\triangle O_1A_1B_1$ . Hence $\phi=\angle OAB=\angle O_1A_1B_1=\phi_1$ .

Same way, $\triangle OAC$ is similar to $\triangle O_1A_1C_1$ . Hence $\theta=\angle OAB=\angle O_1A_1B_1=\theta_1$ .

Combining, $\angle BAC = \varphi + \theta = {\varphi _1} + {\theta _1} = \angle {B_1}{A_1}{C_1}$ .
Now, by SAS (side-angle-side) criterion, $\triangle ABC$ and $\triangle A_1 B_1 C_1$ are similar.

However, beware!
Τhis result is valid only if both circumcenters of the two triangles lay inside the angle subtended by the sides involved in the proportion (or, if both lay outside this angle).

In figure 3 , we see that there exists another triangle, $\triangle A_1B_1C_2$ , that has $A_1B_1=c_1$ , $A_1C_2= A_1C_1=b_1$ and circumradius $R_1$ .
Thus, triangles $\triangle ABC$ and $\triangle A_1 B_1 C_2$ do have two corresponding sides and their circumradii in the same proportion, yet, obviously, they are not similar.

In conclusion, the correct answer is $\boxed{\text{Not enough information}}$ .

Example 1:

$\triangle ABC$ where $a = 1$ , $b = 1$ , and $c = 1$ . Then the area is $K = \frac{\sqrt{3}}{4}$ , and the circumradius is $R = \frac{abc}{4K} = \frac{\sqrt{3}}{3}$ .

$\triangle A'B'C'$ where $a' = 2$ , $b' = 2$ , and $c' = 2$ . Then the area is $K' = \sqrt{3}$ , and the circumradius is $R' = \frac{a'b'c'}{4K'} = \frac{2\sqrt{3}}{3}$ .

Then two of the corresponding sides and the circumradii of these two triangles are in the same proportion ( $\frac{b'}{b} = \frac{c'}{c} = \frac{R'}{R} = 2$ ) and the two triangles are similar.

Example 2:

$\triangle ABC$ where $a = 2$ , $b = 2$ , and $c = 1$ . Then the area (by Heron's formula) is $K = \frac{\sqrt{15}}{4}$ , and the circumradius is $R = \frac{abc}{4K} = \frac{16\sqrt{15}}{15}$ .

$\triangle A'B'C'$ where $a' = 3$ , $b' = 4$ , and $c' = 2$ . Then the area (by Heron's formula) is $K' = \frac{3\sqrt{15}}{4}$ , and the circumradius is $R' = \frac{a'b'c'}{4K'} = \frac{32\sqrt{15}}{15}$ .

Then two of the corresponding sides and the circumradii of these two triangles are in the same proportion ( $\frac{b'}{b} = \frac{c'}{c} = \frac{R'}{R} = 2$ ) but the two triangles are not similar.

The two different examples show that there is not enough information to know if the two triangles are similar.