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, $$\begin{gathered} \frac{AB}{AH}=\frac{AE+EH}{AH}+\frac{HD}{AH}=\frac{1+\cos 2\alpha }{\sin 2\alpha }+\cot \beta \\ =\cot \alpha +\cot \beta .\stackrel{◯}{1} \end{gathered}$$ 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



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