Câu hỏi:

Ta có $\overrightarrow{AM} = \overrightarrow{AC} + \overrightarrow{CM}$ và $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}$.
Vì M là trung điểm BC nên $\overrightarrow{CM} = -\frac{1}{2}\overrightarrow{BC}$.
$\overrightarrow{AM}.\overrightarrow{BC} = (\overrightarrow{AC} - \frac{1}{2}\overrightarrow{BC}).\overrightarrow{BC} = \overrightarrow{AC}.\overrightarrow{BC} - \frac{1}{2}|\overrightarrow{BC}|^2$
$= \overrightarrow{AC}.(\overrightarrow{BA} + \overrightarrow{AC}) - \frac{1}{2}a^2 = \overrightarrow{AC}.\overrightarrow{BA} + |\overrightarrow{AC}|^2 - \frac{1}{2}a^2$
$= b.c.cos(180 - A) + b^2 - \frac{1}{2}a^2 = -bc.cosA + b^2 - \frac{1}{2}a^2$
Theo định lý cosin: $a^2 = b^2 + c^2 - 2bc.cosA \Rightarrow bc.cosA = \frac{b^2 + c^2 - a^2}{2}$
Do đó $\overrightarrow{AM}.\overrightarrow{BC} = -\frac{b^2 + c^2 - a^2}{2} + b^2 - \frac{1}{2}a^2 = \frac{-b^2 - c^2 + a^2 + 2b^2 - a^2}{2} = \frac{b^2 - c^2}{2}$.