证明：延长\(AB\)至点\(E\)，使得\(BE = DN\)，连接\(CE\)，
\(\because \)四边形\(ABCD\)中，\(\angle B + \angle D = 180^\circ \)，\(\angle ABC + \angle CBE = 180^\circ \)，
\(\therefore \angle CBE = \angle CDN\)，
在\(\Delta CBE\)和\(\Delta CDN\)中，
\(\left\{ {\begin{array}{*{20}{l}} {CB = CD} \\\ {\angle CBE = \angle CDN} \\\ {BE = DN} \end{array}} \right.\)，
\(\therefore \Delta CBE \cong \Delta CDN(SAS)\)，
\(\therefore \angle BCE = \angle DCN\)，\(CN = CE\)，
\(\because \angle BCD = 150^\circ \)，\(\angle MCN = 75^\circ \)，
\(\therefore \angle MCE = \angle MCB + \angle BCE = \angle MCB + \angle DCN = 75^\circ \)，
\(\therefore \angle MCN = \angle MCE\)，
在\(\Delta ECM\)和\(\Delta NCM\)中，
\(\left\{ {\begin{array}{*{20}{l}} {MC = MC} \\\ {\angle MCN = \angle MCE} \\\ {CN = CE} \end{array}} \right.\)，
\(\therefore \Delta ECM \cong \Delta NCM(SAS)\)，
\(\therefore MN = ME = BM + BE = BM + DN\)。
