7. As shown in Figure 2, in $\triangle A B C$, $D$ is a point on side $A C$, and $\angle A B D = \angle C$. Point $E$ is on side $A B$, and $B E = D E$. $M$ is the midpoint of side $C D$, and $A H \perp D E$ at point $H$. Given that $A H = 2 - \sqrt{3}$ and $A B = 1$. Find the degree measure of $\angle A M E$.
7. Solution 1 As shown in Figure 4, the circumcircle $\odot O$ of $\triangle BCD$ is tangent to line $AB$ at point $B$.

Draw $AN \perp BD$ at point $N$, and $EG \parallel BD$ intersects $AC$ at point $G$. Connect $OB, OE, OG, OD, OM$.
Since $OE \perp BD, OM \perp CD$, we have
$\angle GEO = 90^\circ = \angle GMO$.
Thus, points $E, G, M, O$ are concyclic.
Therefore, $\tan \angle AME = \tan \angle GOE = \frac{EG}{OE}$.
By $\angle OBE = \angle BNA = 90^\circ$,
$\angle ABN = \angle C = \angle EOB$,
then $\triangle OBE \sim \triangle BNA$.
It is easy to see that points $A, H, N, D$ are concyclic.
Thus, $\angle EDG = \angle ANH$.
Also, $\angle AHN = 180^\circ - \angle ADN = \angle EGD$, so $\triangle EGD \sim \triangle ANH$.
Therefore, $\frac{OE}{BA} = \frac{BE}{NA} = \frac{DE}{NA} = \frac{EG}{AH}$.
Hence, $\frac{EG}{OE} = \frac{AH}{AB} = 2 - \sqrt{3}$.
Thus, $\angle AME = 15^\circ$.
Solution 2 Let $\angle ABD = \angle C = \alpha, \angle DBC = \beta$.
From the given, $\angle BDE = \alpha, \angle AED = 2\alpha$,
$$
\begin{array}{l}
\angle ADE = \angle ADB - \angle BDE \\
= (\alpha + \beta) - \alpha = \beta, \\
AB = AE + EB = AE + EH + HD. \\
\text{Thus, } \frac{AB}{AH} = \frac{AE + EH}{AH} + \frac{HD}{AH} \\
= \frac{1 + \cos 2\alpha}{\sin 2\alpha} + \cot \beta = \cot \alpha + \cot \beta.
\end{array}
$$

As shown in Figure 5, draw $EK \perp AC, EL \perp BD$, with feet at $K, L$ respectively, then $L$ is the midpoint of $BD$.
Using the sine rule, we get
$$
\frac{EL}{EK} = \frac{DE \sin \angle EDL}{DE \sin \angle EDK} = \frac{\sin \alpha}{\sin \beta} = \frac{BD}{CD} = \frac{LD}{MD}.
$$

Then $\cot \alpha = \frac{LD}{EL} = \frac{MD}{EK} = \frac{MK}{EK} - \frac{DK}{EK}$
$$
= \cot \angle AME - \cot \beta.
$$

From equations (1) and (2) and the given conditions, we have
$$
\cot \angle AME = \frac{AB}{AH} = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}.
$$

Thus, $\angle AME = 15^\circ$.