First, call I the intersection of AM and AN.

We have : \(4BN^2=4AB^2+AC^2\)

\(AI^2=AB^2-BI^2=AB^2-\dfrac{4}{9}BN^2\Rightarrow\dfrac{4}{9}AM^2=AB^2-\dfrac{4}{9}\left(AB^2+\dfrac{1}{4}AC^2\right)\)

\(\Rightarrow\dfrac{1}{9}BC^2=\dfrac{5}{9}AB^2-\dfrac{1}{9}AC^2\)

\(\Rightarrow BC^2=5AB^2-AC^2\) and \(BC^2=AB^2+AC^2\Rightarrow AC^2=2AB^2\)

From here we can know that AC = 4.

Good Luck !