Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$. Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$. Tangent form $D$ touches $\Omega$ in $E$. FInd $\angle BEC$. 
1. Identify that point \( D \) is the \( A \)-excenter of \( \triangle ABC \). This means \( D \) is the center of the excircle opposite vertex \( A \).
2. Let \( I \) be the incenter of \( \triangle ABC \), which is the center of the inscribed circle \( \Omega \).
3. Note that the exterior angle bisectors of \( \angle B \) and \( \angle C \) intersect at \( D \), and \( D \) lies on the common side \( BC \).
4. Since \( D \) is the \( A \)-excenter, the angles \( \angle IBD \) and \( \angle ICD \) are both \( 90^\circ \). This is because the incenter and excenter are perpendicular to the sides of the triangle at the points of tangency.
5. Since \( \angle IBD + \angle ICD = 180^\circ \), points \( I, B, C, D \) are concyclic (they lie on the same circle).
6. Given that \( DE \) is tangent to \( \Omega \) at \( E \), we have \( \angle IED = 90^\circ \).
7. Since \( \angle IED = 90^\circ = \angle IBD \), points \( I, E, B, D \) are concyclic.
8. Combining the two sets of concyclic points, we conclude that \( I, E, B, D, C \) are all concyclic.
9. To find \( \angle BEC \), we use the fact that \( \angle BEC = \angle DEB + \angle DEC \).
10. Since \( \angle DEB = \angle DBC \) and \( \angle DEC = \angle DCB \), we have:
    \[
    \angle BEC = \angle DBC + \angle DCB
    \]
11. Using the properties of the excenter and incenter, we know:
    \[
    \angle DBC = 90^\circ - \angle IBC \quad \text{and} \quad \angle DCB = 90^\circ - \angle ICB
    \]
12. Therefore:
    \[
    \angle BEC = (90^\circ - \angle IBC) + (90^\circ - \angle ICB)
    \]
13. Since \( \angle IBC = \frac{\angle ABC}{2} \) and \( \angle ICB = \frac{\angle ACB}{2} \), we have:
    \[
    \angle BEC = 180^\circ - \left( \frac{\angle ABC}{2} + \frac{\angle ACB}{2} \right)
    \]
14. Using the angle sum property of a triangle, \( \angle ABC + \angle ACB = 180^\circ - \angle BAC \), we get:
    \[
    \angle BEC = 180^\circ - \left( \frac{180^\circ - \angle BAC}{2} \right)
    \]
15. Substituting \( \angle BAC = 40^\circ \):
    \[
    \angle BEC = 180^\circ - \left( \frac{180^\circ - 40^\circ}{2} \right) = 180^\circ - \left( \frac{140^\circ}{2} \right) = 180^\circ - 70^\circ = 110^\circ
    \]

The final answer is \(\boxed{110^\circ}\).