A quadrilateral $ABCD$ is inscribed in a circle. Suppose that $|DA| =|BC|= 2$ and$ |AB|  = 4$. Let $E $ be the intersection  point of lines $BC$ and $DA$. Suppose that $\angle AEB = 60^o$ and that $|CD| <|AB|$. Calculate the radius of the circle.
1. Given that quadrilateral $ABCD$ is inscribed in a circle, we know that $|DA| = |BC| = 2$ and $|AB| = 4$. Let $E$ be the intersection point of lines $BC$ and $DA$. We are also given that $\angle AEB = 60^\circ$ and $|CD| < |AB|$.

2. Since $ABCD$ is a cyclic quadrilateral, we can use the property that opposite angles of a cyclic quadrilateral sum to $180^\circ$. However, in this problem, we will focus on the triangle $\triangle AEB$.

3. Notice that $\angle AEB = 60^\circ$. This suggests that $\triangle AEB$ might be an equilateral triangle because all angles in an equilateral triangle are $60^\circ$.

4. If $\triangle AEB$ is equilateral, then $|AE| = |EB| = |AB| = 4$. This is because all sides of an equilateral triangle are equal.

5. Since $D$ is on $AE$ and $|DA| = 2$, $D$ is the midpoint of $AE$. Therefore, $|DE| = |DA| = 2$.

6. Now, consider the circle in which quadrilateral $ABCD$ is inscribed. The diameter of the circle can be found using the fact that $AB$ is a chord of the circle and $\angle AEB = 60^\circ$.

7. In an equilateral triangle inscribed in a circle, the side length is equal to the radius of the circle times $\sqrt{3}$. However, since $AB = 4$ is the side of the equilateral triangle, we can use the property that the radius of the circumcircle of an equilateral triangle is $\frac{\text{side length}}{\sqrt{3}}$.

8. Therefore, the radius $R$ of the circle is:
   \[
   R = \frac{AB}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}
   \]

9. However, since $AB = 4$ is the diameter of the circle (as $\angle AEB = 60^\circ$ and $D$ is the midpoint of $AE$), the radius is half of the diameter:
   \[
   R = \frac{AB}{2} = \frac{4}{2} = 2
   \]

The final answer is $\boxed{2}$.