A circle of radius $2$ is inscribed in equilateral triangle $ABC$. The altitude from $A$ to $BC$ intersects the circle at a point $D$ not on $BC$. $BD$ intersects the circle at a point $E$ distinct from $D$. Find the length of $BE$.
1. **Identify the given information and draw the diagram:**
   - A circle of radius \(2\) is inscribed in an equilateral triangle \(ABC\).
   - The altitude from \(A\) to \(BC\) intersects the circle at point \(D\) not on \(BC\).
   - \(BD\) intersects the circle at point \(E\) distinct from \(D\).

2. **Determine the side length of the equilateral triangle:**
   - Let the side length of the equilateral triangle be \(s\).
   - The altitude of an equilateral triangle splits it into two 30-60-90 right triangles.
   - The altitude \(h\) can be expressed as \(h = \frac{\sqrt{3}}{2} s\).

3. **Find the relationship between the radius of the inscribed circle and the side length:**
   - The radius \(r\) of the inscribed circle in an equilateral triangle is given by \(r = \frac{s \sqrt{3}}{6}\).
   - Given \(r = 2\), we have:
     \[
     2 = \frac{s \sqrt{3}}{6} \implies s = \frac{12}{\sqrt{3}} = 4\sqrt{3}
     \]

4. **Locate the incenter \(O\) and its distance from the vertices:**
   - The incenter \(O\) is also the centroid of the equilateral triangle.
   - The distance from the centroid to any vertex is \(\frac{2}{3}\) of the altitude.
   - Therefore, \(BO = \frac{2}{3} \times \frac{\sqrt{3}}{2} s = \frac{\sqrt{3}}{3} s = \frac{\sqrt{3}}{3} \times 4\sqrt{3} = 4\).

5. **Calculate \(BD\):**
   - Since \(D\) is on the circle and \(O\) is the center, \(OD = 2\).
   - \(BD\) is the hypotenuse of the right triangle \(BOD\) with \(BO = 4\) and \(OD = 2\):
     \[
     BD = \sqrt{BO^2 - OD^2} = \sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}
     \]

6. **Find \(BE\):**
   - Since \(E\) is the other intersection point of \(BD\) with the circle, \(DE\) is a chord of the circle.
   - The length of the chord \(DE\) can be found using the fact that \(D\) and \(E\) are symmetric with respect to the diameter through \(B\).
   - Therefore, \(BE = BD = 2\sqrt{3}\).

The final answer is \(\boxed{2\sqrt{3}}\).