On the extension of chord $AB$ of a circle centroid at $O$ a point $X$ is taken and tangents $XC$ and $XD$ to the circle are drawn from it with $C$ and $D$ lying on the circle, let $E$ be the midpoint of the line segment $CD$. If $\angle OEB = 140^o$ then determine with proof the magnitude of $\angle AOB$.
1. **Identify the given elements and relationships:**
   - \( O \) is the center of the circle.
   - \( AB \) is a chord of the circle.
   - \( X \) is a point on the extension of \( AB \).
   - \( XC \) and \( XD \) are tangents to the circle from \( X \), with \( C \) and \( D \) lying on the circle.
   - \( E \) is the midpoint of \( CD \).
   - \(\angle OEB = 140^\circ\).

2. **Establish the collinearity of \( O, E, \) and \( X \):**
   - Since \( E \) is the midpoint of \( CD \) and \( O \) is the center of the circle, \( OE \) is the perpendicular bisector of \( CD \). Therefore, \( O, E, \) and \( X \) are collinear.

3. **Use the Power of a Point theorem:**
   - By the Power of a Point theorem, for point \( X \) outside the circle, the power of \( X \) is equal for both tangents and secants:
     \[
     XC^2 = XO \cdot XE
     \]
     and
     \[
     XC^2 = XB \cdot XA
     \]
   - Equating these, we get:
     \[
     XO \cdot XE = XB \cdot XA
     \]

4. **Conclude that points \( B, A, O, E \) are concyclic:**
   - Since \( XO \cdot XE = XB \cdot XA \), points \( B, A, O, E \) lie on a common circle.

5. **Determine \(\angle OAB\):**
   - Since \( B, A, O, E \) are concyclic, the opposite angles of the cyclic quadrilateral sum to \( 180^\circ \):
     \[
     \angle OAB + \angle OEB = 180^\circ
     \]
   - Given \(\angle OEB = 140^\circ\), we have:
     \[
     \angle OAB = 180^\circ - 140^\circ = 40^\circ
     \]

6. **Calculate \(\angle AOB\):**
   - The central angle \(\angle AOB\) is twice the inscribed angle \(\angle OAB\):
     \[
     \angle AOB = 2 \times \angle OAB = 2 \times 40^\circ = 80^\circ
     \]

The final answer is \(\boxed{80^\circ}\).