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Denote $\beta=\angle MBC,\;\alpha=\angle DMN,\;\gamma=\angle CDH$. We have: $$\tan\beta=1/2,\quad\cos\alpha=-\sin\beta=-1/\sqrt{5},\quad\sin\alpha=\cos\beta=2/\sqrt{5}.$$ Take $DM=1$, then $MN=\sqrt{5}-2$, and let $x=DN$. By the rule of cosines, $$x^2=1^2+(\sqrt{5}-2)^2+2(\sqrt{5}-2)/\sqrt{5}=12-24/\sqrt{5}.$$ Then by the rule of sines, \sin\gamma=\frac{(\...