Ta có: \(\dfrac{AB}{AC}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{HB}{HC}=\dfrac{1}{16}\)
hay HC=16HB
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow16HB^2=148\)
\(\Leftrightarrow HB=\dfrac{\sqrt{37}}{2}\)
\(\Leftrightarrow HC=8\sqrt{37}\)
\(\Leftrightarrow BC=\dfrac{17\sqrt{37}}{2}\left(cm\right)\)