\[{S_{ABC}} = \frac{1}{2} \cdot AC \cdot BC \cdot \sin \widehat {ACB}\]

\[{S_{CMN}} = \frac{1}{2} \cdot CM \cdot CN.\sin \widehat {ACB} = \frac{1}{2} \cdot \frac{1}{2}AC \cdot \frac{1}{2}BC \cdot \sin \widehat {ACB}\]

Nên \({S_{CMN}} = \frac{1}{4}{S_{ABC}}\)

\( \Rightarrow {S_{ABC}} = 4{S_{CMN}} = 60\sqrt 3 \)

\({S_{ABG}} = \frac{2}{3}{S_{ABM}} = \frac{2}{3} \cdot \frac{1}{2}{S_{ABC}} = \frac{1}{3} \cdot 60\sqrt 3 = 20\sqrt 3 \)

Lại có: \({S_{ABG}} = \frac{1}{2} \cdot AG \cdot GB \cdot {\rm{sin}}\widehat {AGB}\)

                     \( = \frac{1}{2} \cdot \frac{2}{3}AM \cdot \frac{2}{3}BN \cdot {\rm{sin}}\widehat {AGB} = 40 \cdot {\rm{sin}}\widehat {AGB}\)

Suy ra \(40 \cdot {\rm{sin}}\widehat {AGB} = 20\sqrt 3 \)

\( \Rightarrow {\rm{sin}}\widehat {AGB} = \frac{{\sqrt 3 }}{2}\)

\( \Rightarrow \widehat {AGB} = 60^\circ \)

Ta có: \(A{B^2} = A{G^2} + G{B^2} - 2 \cdot AG \cdot GB \cdot {\rm{cos}}\widehat {AGB}\)

\( = {\left( {\frac{2}{3} \cdot 15} \right)^2} + {\left( {\frac{2}{3} \cdot 12} \right)^2} - 2 \cdot \frac{2}{3}15 \cdot \frac{2}{3} \cdot 12 \cdot {\rm{cos}}60^\circ = 84\)

\( \Rightarrow AB = 2\sqrt {21} \)

M, N là trung điểm của BC, AC

⇒ MN là đường trung bình của ∆ACB

\( \Rightarrow MN = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 2\sqrt {21} = \sqrt {21} \)

Vậy \(MN = \sqrt {21} .\)