China Southeastern Mathematical Olympiad

Let $\angle ABD = \angle C = \alpha$ and $\angle DBC = \beta$. It is easy to see that $\angle BDE = \alpha$, $\angle AED = 2\alpha$,
$$\angle ADE = \angle ADB - \angle BDE = (\alpha + \beta) - \alpha = \beta,$$
$$AB = AE + EB = AE + EH + HD.$$
Hence,
$$\frac{AB}{AH} = \frac{AE + EH}{AH} + \frac{HD}{AH} = \frac{1 + \cos 2\alpha}{\sin 2\alpha} + \cot \beta \\ = \cot \alpha + \cot \beta. \qquad \textcircled{1}$$
Draw lines $EK \perp AC$ and $EL \perp BD$ with pedals $K$ and $L$, respectively. Then, $L$ is the midpoint of $BD$. Combining with the Sine Theorem, we obtain

$$\frac{EL}{EK} = \frac{DE \sin \angle EDL}{DE \sin \angle EDK} = \frac{\sin \alpha}{\sin \beta} \\ = \frac{BD}{CD} = \frac{LD}{MD}.$$
Thus,
$$\cot \alpha = \frac{LD}{EL} = \frac{MD}{EK} = \frac{MK}{EK} - \frac{DK}{EK} = \cot \angle AME - \cot \beta. \quad \textcircled{2}$$
By $\textcircled{1}$, $\textcircled{2}$ and known conditions, we have
$$\cot \angle AME = \frac{AB}{AH} = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}.$$
Therefore, $\angle AME = 15^\circ$.