Let the circumradius and inradius of $\triangle A B C$ be $R, r$, respectively. Try to express $\frac{M N}{B C}$ using $R, r$.
Let $O, I$ be the circumcenter and incenter of $\triangle ABC$, respectively. Connect $BI$ and extend it to intersect the circumcircle of $\triangle ABC$ at $D$, intersect $MC$ at point $E$, and intersect $AC$ at point $F$. Connect $AD, CD, OD$, and let $OD$ intersect $AC$ at $G$. It is easy to see that $AD = CD, OD \perp AC$, so $\angle ODI = \angle ADI - \angle ADG = \angle ACB - (90^\circ - \angle CAD) = \angle ACB - 90^\circ + \angle DBC = \angle ACE$ (since $BM = BC$, so $BE \perp CE$).

It is also easy to see that $DI = CD$, so $\frac{CN}{OD} = \frac{BC}{OD} = 2 \sin A = 2 \sin \angle BDC = 2 \frac{CE}{CD} = \frac{CM}{DI}$, thus $\triangle CMN \sim \triangle DIO$, so $\frac{MN}{OI} = \frac{CM}{DI} = 2 \sin A$, so $MN = 2 OI \sin A$. By a well-known result, $OI^2 = R^2 - 2Rr$, so $\frac{MN}{BC} = \frac{2 OI \sin A}{BC} = \frac{OI}{R} = \frac{\sqrt{R^2 - 2Rr}}{R} = \sqrt{1 - \frac{2r}{R}}$.