Let $ABCDEF$ be a regular hexagon and $M\in (DE)$, $N\in(CD)$ such that $m (\widehat {AMN}) = 90^\circ$ and $AN = CM \sqrt {2}$. Find the value of $\frac{DM}{ME}$.
Since $\widehat{AMN}=\widehat{ACN}=90^{\circ},$ then $AN$ is a diameter of the circumcircle of $\triangle ACM$ $\Longrightarrow$ $CM=AN \cdot \sin \widehat{MAC}$ $\Longrightarrow$ $\sin \widehat{MAC}=\frac{_1}{^{\sqrt{2}}}$ $\Longrightarrow$ $\widehat{MAC}=45^{\circ},$ since $\widehat{MAC} < 60^{\circ}.$ Thus $\widehat{EAM}=60^{\circ}-45^{\circ}=15^{\circ}$ $\Longrightarrow$ $AM$ bisects $\widehat{DAE}.$ By angle bisector theorem on $\triangle DAE,$ we have then
$\frac{DM}{ME}=\frac{DA}{AE}=\frac{\frac{2\sqrt{3}}{3}AE}{AE}=\frac{2\sqrt{3}}{3}.$