Points $E$ and $C$ are chosen on a semicircle with diameter $AB$ and center $O$ such that $OE \perp AB$ and the intersection point $D$ of $AC$ and $OE$ is inside the semicircle.  Find all values of $\angle{CAB}$ for which the quadrilateral $OBCD$ is tangent.
1. **Identify the given elements and their relationships:**
   - Points \( E \) and \( C \) are on a semicircle with diameter \( AB \) and center \( O \).
   - \( OE \perp AB \).
   - \( D \) is the intersection point of \( AC \) and \( OE \) inside the semicircle.
   - Quadrilateral \( OBCD \) is tangent.

2. **Define the center of the tangential quadrilateral:**
   - Let \( W \) be the center of the circle inscribed in quadrilateral \( OBCD \).

3. **Prove that \( D, W, B \) are collinear:**
   - Since \( W \) is the center of the inscribed circle, it lies on the angle bisectors of \( \angle OBD \) and \( \angle ODC \).
   - Therefore, \( W \) must lie on the line \( DB \).

4. **Determine the angle relationships:**
   - Let \( \alpha \) be the angle \( \angle OBW \).
   - Since \( W \) is the incenter, \( \angle OBW = \frac{\angle OBD}{2} \).

5. **Use tangent properties to find \( \alpha \):**
   - The tangent of the angle \( \angle ODB \) can be expressed as:
     \[
     \tan(\angle ODB) = \frac{OB}{OD} = \tan(2\alpha)
     \]
   - The tangent of the angle \( \angle ADO \) can be expressed as:
     \[
     \tan(\angle ADO) = \frac{OD}{AO} = \frac{OD}{OB} = \tan(45^\circ + \alpha)
     \]

6. **Set up the equation involving \( \alpha \):**
   - From the tangent properties, we have:
     \[
     \tan(2\alpha) = \frac{1}{\tan(45^\circ + \alpha)}
     \]
   - Using the identity \( \tan(45^\circ + \alpha) = \frac{1 - \tan(\alpha)}{1 + \tan(\alpha)} \), we get:
     \[
     \cot(2\alpha) = \tan(45^\circ + \alpha)
     \]

7. **Solve the equation for \( \alpha \):**
   - The equation \( \cot(2\alpha) = \tan(45^\circ + \alpha) \) simplifies to:
     \[
     45^\circ + \alpha + 2\alpha = 90^\circ \implies 3\alpha = 45^\circ \implies \alpha = 15^\circ
     \]
   - Therefore, \( \frac{\angle CAB}{2} = 15^\circ \) which implies:
     \[
     \angle CAB = 30^\circ
     \]

The final answer is \(\boxed{30^\circ}\)