\(\overrightarrow {BM} = \frac{1}{3}\overrightarrow {MC} \Rightarrow \overrightarrow {BM} = \frac{1}{4}\overrightarrow {BC} \)

\(\overrightarrow {CN} = k\overrightarrow {AN} \Rightarrow \overrightarrow {CN} = - k\overrightarrow {NA} \Rightarrow \overrightarrow {CA} = \left( {1 - k} \right)\overrightarrow {NA}  \Rightarrow \overrightarrow {AN} = \frac{1}\overrightarrow {AC} \)

AM vuông góc với DN khi và chỉ khi: \(\overrightarrow {AM} .\overrightarrow {DN} = 0\)

⇔ \(\left( {\overrightarrow {AB} + \overrightarrow {BM} } \right).\left( {\overrightarrow {DA} + \overrightarrow {AN} } \right) = 0\)

⇔\(\left( {\overrightarrow {AB} + \frac{1}{4}\overrightarrow {BC} } \right).\left( {\overrightarrow {DC} + \frac{1}\overrightarrow {AC} } \right) = 0\)

⇔\[\overrightarrow {AB} .\overrightarrow {DC} + \frac{1}.\overrightarrow {AB} .\overrightarrow {DC} + \frac{1}{4}\overrightarrow {BC} .\overrightarrow {DC} + \frac{1}{4}.\frac{1}.\overrightarrow {BC} .\overrightarrow {AC} = 0\]

⇔ \[a.a.\cos 0^\circ + \frac{1}.a.a\sqrt 2 .\cos 45^\circ + \frac{1}{4}.a.a.\cos 90^\circ + \frac{1}{4}.\frac{1}.a.a\sqrt 2 .\cos 45^\circ = 0\]

⇔\[{a^2} + \frac{{{a^2}}} + \frac{{{a^2}}}{{4\left( {1 - k} \right)}} = 0\]

⇔ \[{a^2}\left( {1 + \frac{1} + \frac{1}{{4\left( {1 - k} \right)}}} \right) = 0\]

⇔ \[\frac{{4\left( {1 - k} \right)}} = 0\]

⇔ –4k + 9 = 0

⇔ k = \(\frac{9}{4}\).