Draw \(ND\perp AB\left(D\in AB\right)\) ,then ND // AC since \(AB\perp AC\)

\(\Rightarrow\dfrac{BD}{AB}=\dfrac{ND}{AC}=\dfrac{BN}{BC}=\dfrac{1}{3}\)(Intercept theorem of Thales)

Applying the Pythagoras theorem to \(\Delta ABM,\Delta ADN\),we have :

\(\dfrac{BM^2}{AN^2}=\dfrac{AB^2+AM^2}{AD^2+ND^2}=\dfrac{AB^2+\left(\dfrac{AC}{2}\right)^2}{\left(\dfrac{2}{3}AB\right)^2+\left(\dfrac{AC}{3}\right)^2}\)

\(=\dfrac{AB^2+\dfrac{1}{4}AC^2}{\dfrac{4}{9}AB^2+\dfrac{1}{9}AC^2}=\dfrac{\dfrac{1}{4}\left(4AB^2+AC^2\right)}{\dfrac{1}{9}\left(4AB^2+AC^2\right)}=\dfrac{9}{4}\)

\(\Rightarrow BM^2=\dfrac{9}{4}AN^2\Rightarrow BM=\dfrac{3}{2}AN=\dfrac{3}{2}.34=51\left(cm\right)\)