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 $\angle AME$.
Since $\angle ABD=\angle EDB=\angle ACB,$ then $EB,ED$ are tangent to the circumcircle $(K)$ of $\triangle BCD.$ Let $\odot(KME)$ cut $AC$ again at $F.$ $\angle FEK=\angle FMK=90^{\circ}$ $\Longrightarrow$ $EF \parallel BD \perp EK$ $\Longrightarrow$ $\tfrac{EF}{BD}=\tfrac{AE}{AB}.$ Moreover $\tfrac{AH}{AE}=\sin \widehat{AED}=\sin \widehat{BKD}=\tfrac{BD}{EK}.$ Hence
$\tan \widehat{AME}=\tan \widehat{FKE}=\frac{EF}{EK}=\frac{BD}{EK} \cdot \frac{AE}{AB}= \frac{AH}{AE} \cdot \frac{AE}{AB}=\frac{AH}{AB}=2-\sqrt{3}$
$\Longrightarrow \widehat{AME}=\arctan (2-\sqrt{3})=15^{\circ}.$
P.S. See the topic the degree of AME for another solution.