Vì $\widehat{A} = 90^\circ$ nên $\overrightarrow{AB}\cdot\overrightarrow{AC} = 0$, $|\overrightarrow{AB}|^2 = c^2 = 36$, $|\overrightarrow{AC}|^2 = b^2 = 64.$
Vì $BM = 3\,MC$ nên $M$ chia $BC$ theo tỉ số $\dfrac{BM}{MC} = 3$, do đó $\overrightarrow{AM} = \dfrac{1}{1+3}\,\overrightarrow{AB} + \dfrac{3}{1+3}\,\overrightarrow{AC} = \dfrac{1}{4}\overrightarrow{AB} + \dfrac{3}{4}\overrightarrow{AC}.$
Mặt khác $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}.$
$\overrightarrow{AM}\cdot\overrightarrow{BC} = \left[\dfrac{1}{4}\overrightarrow{AB} + \dfrac{3}{4}\overrightarrow{AC}\right]\cdot(\overrightarrow{AC} - \overrightarrow{AB}) = -\dfrac{1}{4}|\overrightarrow{AB}|^2 + \dfrac{3}{4}|\overrightarrow{AC}|^2 + 0 = \dfrac{3\cdot 64 - 36}{1+3} = 39.$