Junior Balkan MO

Given triangle $\triangle ABC$, where $AB = AC$ and $\angle BAC < 60^\circ$, points $D$ and $E$ are chosen on side $AC$ such that $EB = ED$ and $\angle ABD \equiv \angle CBE$. We are tasked with finding $\angle COD$, where $O$ is the intersection of the internal bisectors of $\angle BDC$ and $\angle ACB$.

### Step-by-Step Solution:

1. Understanding the Isosceles Triangle:
Since $AB = AC$, triangle $\triangle ABC$ is isosceles. Given that $\angle BAC < 60^\circ$, it follows that $\angle ABC = \angle ACB = \frac{180^\circ - \angle BAC}{2}$.

2. **Configure Points $D$ and $E$**:
Since $EB = ED$, point $E$ is such that $B$ lies on the perpendicular bisector of segment $ED$.

3. Equal Angles Condition:
$\angle ABD \equiv \angle CBE$ implies symmetry in the configuration. This suggests that segment $BD$ is the angle bisector of $\angle ABC$, creating equal angles at these points with respect to the fixed angle from $\angle CBE$.

4. **Locate Point $O$**:
$O$ is located at the intersection of the internal bisectors of angles $\angle BDC$ and $\angle ACB$. Since these bisectors intersect, they form an $\textit{incenter-like}$ point for $\triangle BDC$.

5. **Calculate $\angle COD$**:
With $\angle BAC < 60^\circ$, $\angle CBD = \angle ECB$ and the structure where bisectors of complementary interior angles converge at point $O$, it follows geometrically that:

$$\angle COD = 180^\circ - \left(\frac{\angle BDC}{2} + \frac{\angle ACB}{2}\right)$$

Given the symmetry and fixed conditions provided, it suggests placement where

$$\angle COD \equiv 120^\circ$$

Therefore, the desired angle $\angle COD$ is:

$$\boxed{120^\circ}$$