In $\triangle ABC$, point $D$ lies on side $AC$ such that $\angle ABD=\angle C$. Point $E$ lies on side $AB$ such that $BE=DE$. $M$ is the midpoint of segment $CD$. Point $H$ is the foot of the perpendicular from $A$ to $DE$. Given $AH=2-\sqrt{3}$ and $AB=1$, find the size of $\angle AME$.
1. **Given Information and Initial Setup:**
   - In $\triangle ABC$, point $D$ lies on side $AC$ such that $\angle ABD = \angle C$.
   - Point $E$ lies on side $AB$ such that $BE = DE$.
   - $M$ is the midpoint of segment $CD$.
   - Point $H$ is the foot of the perpendicular from $A$ to $DE$.
   - Given $AH = 2 - \sqrt{3}$ and $AB = 1$.

2. **Angles and Tangency:**
   - Since $\angle ABD = \angle C$, it follows that $\angle ABD = \angle EDB = \angle ACB$.
   - This implies that $EB$ and $ED$ are tangent to the circumcircle $(K)$ of $\triangle BCD$.

3. **Constructing Circle and Parallel Lines:**
   - Let the circle $\odot(KME)$ intersect $AC$ again at $F$.
   - Since $\angle FEK = \angle FMK = 90^\circ$, it follows that $EF \parallel BD \perp EK$.

4. **Proportionality and Trigonometric Relationships:**
   - From the parallel lines, we have $\frac{EF}{BD} = \frac{AE}{AB}$.
   - Additionally, $\frac{AH}{AE} = \sin \angle AED = \sin \angle BKD = \frac{BD}{EK}$.

5. **Calculating $\tan \angle AME$:**
   - Using the above relationships, we get:
     \[
     \tan \angle AME = \tan \angle FKE = \frac{EF}{EK} = \frac{BD}{EK} \cdot \frac{AE}{AB} = \frac{AH}{AE} \cdot \frac{AE}{AB} = \frac{AH}{AB}
     \]
   - Substituting the given values, we find:
     \[
     \frac{AH}{AB} = 2 - \sqrt{3}
     \]
   - Therefore:
     \[
     \tan \angle AME = 2 - \sqrt{3}
     \]

6. **Finding the Angle:**
   - We need to find the angle whose tangent is $2 - \sqrt{3}$.
   - We recognize that $\tan 15^\circ = 2 - \sqrt{3}$.

7. **Conclusion:**
   - Thus, $\angle AME = 15^\circ$.

The final answer is $\boxed{15^\circ}$.