Đặt \(AC=x\Rightarrow BC^2=a^2+x^2\)
\(AM=\frac{1}{2}BC\Rightarrow AM^2=\frac{1}{4}BC^2=\frac{a^2+x^2}{4}\)
\(AG=\frac{2}{3}AM\Rightarrow AG^2=\frac{4}{9}AM^2=\frac{a^2+x^2}{9}\)
\(BG=\frac{2}{3}BN\Rightarrow BG^2=\frac{4}{9}BN^2=\frac{4}{9}\left(AB^2+AN^2\right)\)
\(\Rightarrow BG^2=\frac{4}{9}\left(a^2+\frac{x^2}{4}\right)=\frac{4a^2+x^2}{9}\)
Pitago tam giác ABG:
\(BG^2+AG^2=AB^2\Leftrightarrow\frac{4a^2+x^2}{9}+\frac{a^2+x^2}{9}=a^2\)
\(\Leftrightarrow5a^2+2x^2=9a^2\Rightarrow x^2=2a^2\)
\(\Rightarrow BN=\sqrt{AB^2+AN^2}=\sqrt{a^2+\frac{x^2}{4}}=\sqrt{a^2+\frac{a^2}{2}}=\frac{a\sqrt{6}}{2}\)