From the vertex $ A$ of the equilateral triangle $ ABC$ a line is drown that intercepts the segment $ [BC]$ in the point $ E$. The point $ M \in (AE$ is such that $ M$ external to $ ABC$, $ \angle AMB \equal{} 20 ^\circ$ and $ \angle AMC \equal{} 30 ^ \circ$. What is the measure of the angle $ \angle MAB$?
1. **Given:**
   - $ABC$ is an equilateral triangle.
   - $M$ is a point such that $M$ is external to $ABC$.
   - $\angle AMB = 20^\circ$ and $\angle AMC = 30^\circ$.
   - We need to find the measure of $\angle MAB$.

2. **Case 1: $E \in (BC)$**
   - Denote $m(\angle BAM) = x$.
   - Since $ABC$ is equilateral, $m(\angle BAC) = 60^\circ$.
   - Therefore, $m(\angle MAC) = 60^\circ - x$.
   - Since $M$ is external to $ABC$, $m(\angle BCM) = 30^\circ + x$ and $m(\angle MBC) = 100^\circ - x$.

3. **Using the Sine Rule in Quadrilateral $ABMC$:**
   \[
   \sin(\angle BAM) \cdot \sin(\angle MBC) \cdot \sin(\angle CMA) \cdot \sin(\angle ACB) = \sin(\angle MAC) \cdot \sin(\angle CBA) \cdot \sin(\angle AMB) \cdot \sin(\angle BCM)
   \]
   Substituting the known values:
   \[
   \sin(x) \cdot \sin(100^\circ - x) \cdot \sin(30^\circ) \cdot \sin(60^\circ) = \sin(60^\circ - x) \cdot \sin(60^\circ) \cdot \sin(20^\circ) \cdot \sin(30^\circ + x)
   \]
   Simplifying:
   \[
   \sin(x) \cdot \sin(100^\circ - x) \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \sin(60^\circ - x) \cdot \frac{\sqrt{3}}{2} \cdot \sin(20^\circ) \cdot \sin(30^\circ + x)
   \]
   \[
   \sin(x) \cdot \sin(100^\circ - x) = 2 \cos(30^\circ + x) \cdot \sin(20^\circ) \cdot \sin(30^\circ + x)
   \]
   \[
   \sin(x) \cdot \cos(10^\circ - x) = \sin(20^\circ) \cdot \cos(30^\circ - 2x)
   \]
   \[
   \sin(10^\circ) + \sin(2x - 10^\circ) = \sin(50^\circ - 2x) + \sin(2x - 10^\circ)
   \]
   \[
   \sin(10^\circ) = \sin(50^\circ - 2x)
   \]
   Solving for $x$:
   \[
   10^\circ = 50^\circ - 2x \quad \text{or} \quad 10^\circ + (50^\circ - 2x) = 180^\circ
   \]
   \[
   x = 20^\circ \quad \text{or} \quad x = -60^\circ \not\in [0^\circ, 180^\circ]
   \]
   Therefore, $x = 20^\circ$.

4. **Case 2: $B \in (EC)$**
   - Denote $m(\angle BAM) = x$.
   - Since $M$ is external to $ABC$, $m(\angle BCM) = 30^\circ - x$ and $m(\angle BMC) = 10^\circ$.

5. **Using the Trigonometric Form of Ceva's Theorem:**
   \[
   \sin(\angle BAM) \cdot \sin(\angle BMC) \cdot \sin(\angle BCA) = \sin(\angle BAC) \cdot \sin(\angle BMA) \cdot \sin(\angle BCM)
   \]
   Substituting the known values:
   \[
   \sin(x) \cdot \sin(10^\circ) \cdot \sin(60^\circ) = \sin(60^\circ) \cdot \sin(20^\circ) \cdot \sin(30^\circ - x)
   \]
   Simplifying:
   \[
   \sin(x) \cdot \sin(10^\circ) = \sin(20^\circ) \cdot \sin(30^\circ - x)
   \]
   \[
   \cos(x - 10^\circ) - \cos(x + 10^\circ) = \cos(x - 10^\circ) - \cos(50^\circ - x)
   \]
   \[
   \cos(x + 10^\circ) = \cos(50^\circ - x)
   \]
   Solving for $x$:
   \[
   x + 10^\circ = 50^\circ - x
   \]
   \[
   x = 20^\circ
   \]

Conclusion:
In both cases, we find that $m(\angle MAB) = 20^\circ$.

The final answer is $\boxed{20^\circ}$