Cho tam giác ABC có \(BC =...

D

- Hướng dẫn giải

Ta có:

$\overrightarrow {AM} .\overrightarrow {BC} = \frac{1}{2}(\overrightarrow {AB} + \overrightarrow {AC} ).\overrightarrow {BC} = \frac{1}{2}(\overrightarrow {AB} .\overrightarrow {BC} + \overrightarrow {AC} .\overrightarrow {BC} )$

$= \frac{1}{2}{\rm{[}}c.a.cos({180^0} - B) + b.a.cos({180^0} - C){\rm{]}}$

$= - (c.a.cosB + b.a.cosC)$

$= - (c.a.\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} + ab.\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}})$

$= - \frac{1}{2}.2{a^2} = - {a^2}$