In $\triangle ABC$ points $M,O$ are midpoint of $AB$ and circumcenter. It is true, that $OM=R-r$. Bisector of external $\angle A$ intersect $BC$ at $D$ and bisector of external $\angle C$ intersect $AB$ at $E$. 
Find possible values of $\angle CED$

[i]D. Shiryaev [/i]
1. **Given**: In $\triangle ABC$, points $M$ and $O$ are the midpoint of $AB$ and the circumcenter, respectively. It is given that $OM = R - r$, where $R$ is the circumradius and $r$ is the inradius. The external angle bisector of $\angle A$ intersects $BC$ at $D$, and the external angle bisector of $\angle C$ intersects $AB$ at $E$.

2. **Objective**: Find the possible values of $\angle CED$.

3. **Step-by-Step Solution**:
   - Let $AB = c$, $BC = a$, and $CA = b$.
   - Since $M$ is the midpoint of $AB$, $M$ divides $AB$ into two equal segments, each of length $\frac{c}{2}$.
   - Given $OM = R - r$, we use the distance formula in the context of the circumcenter and the midpoint of a side of the triangle.

4. **Using the given condition**:
   \[
   OM^2 = (R - r)^2
   \]
   - The distance $OM$ can also be expressed using the coordinates of $O$ and $M$. Since $M$ is the midpoint of $AB$, we have:
   \[
   OM^2 = R^2 - \left(\frac{c}{2}\right)^2
   \]
   - Equating the two expressions for $OM^2$:
   \[
   (R - r)^2 = R^2 - \left(\frac{c}{2}\right)^2
   \]
   - Expanding and simplifying:
   \[
   R^2 - 2Rr + r^2 = R^2 - \frac{c^2}{4}
   \]
   - Canceling $R^2$ from both sides:
   \[
   -2Rr + r^2 = -\frac{c^2}{4}
   \]
   - Rearranging:
   \[
   2Rr - r^2 = \frac{c^2}{4}
   \]

5. **Solving for $c$**:
   - From the above equation, we can solve for $c$:
   \[
   c^2 = 4(2Rr - r^2)
   \]
   - This implies:
   \[
   c = \sqrt{4(2Rr - r^2)}
   \]

6. **Considering the angle bisectors**:
   - Let the internal bisectors of $\angle BAC$ and $\angle ACB$ intersect the opposite sides at $F$ and $G$, respectively.
   - Using the angle bisector theorem, we have:
   \[
   \frac{BE}{BG} = \frac{BD}{DC}
   \]
   - This implies that $DE \parallel CG$.

7. **Finding $\angle CED$**:
   - Since $DE \parallel CG$, the angle $\angle CED$ is equal to $\angle ECG$.
   - Given that $CG$ is a line segment and $DE \parallel CG$, $\angle ECG$ is a right angle:
   \[
   \angle CED = \angle ECG = 90^\circ
   \]

The final answer is $\boxed{90^\circ}$.