Cho tam giác ABC có BC = a, CA = b, AB = c. Trên cạnh AB lấy điểm M. Khẳng định nào

Hướng dẫn giải:

Đáp án đúng là: A.

Do M thuộc cạnh AB nên ta có:

$\overrightarrow{AM}=-\frac{AM}{BM}.\overrightarrow{BM}$

$\iff \left(\overrightarrow{AC}+\overrightarrow{CM}\right)=-\frac{AM}{BM}.\left(\overrightarrow{BC}+\overrightarrow{CM}\right)$

$\iff \left(1+\frac{AM}{BM}\right)\overrightarrow{CM}=-\frac{AM}{BM}\overrightarrow{BC}-\overrightarrow{AC}$

$\iff \frac{BM+AM}{BM}\overrightarrow{CM}=\frac{AM}{BM}\overrightarrow{CB}+\overrightarrow{CA}$

$\iff \frac{AB}{BM}\overrightarrow{CM}=\frac{AM}{BM}\overrightarrow{CB}+\frac{BM}{BM}\overrightarrow{CA}$

$\iff AB.\overrightarrow{CM}=AM.\overrightarrow{CB}+\overrightarrow{BM}.\overrightarrow{CA}$

$\iff c.\overrightarrow{CM}=AM.\overrightarrow{CB}+BM.\overrightarrow{CA}$

$\iff {\left(c.\overrightarrow{CM}\right)}^2={\left(AM.\overrightarrow{CB}+BM.\overrightarrow{CA}\right)}^2$

$\iff c^2.C{M}^2=A{M}^2.C{B}^2+B{M}^2.C{A}^2+2AM.BM.\overrightarrow{CA}.\overrightarrow{CB}$

$\iff c^2.C{M}^2=a^2.A{M}^2+b^2.B{M}^2+2AM.BM.CA.CB.\cos C$

 $\iff c^2.C{M}^2=a^2.A{M}^2+b^2.B{M}^2+2AM.BM.b.a.\frac{a^2+b^2-c^2}{2ab}$  (định lí côsin)