Cho tam giác ABC. Các điểm D, E thuộc cạnh BC thỏa mãn BD = DE = EC (Hình 62)

+ Ta có: $\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-\overrightarrow{AB}+\overrightarrow{AC}=-\overrightarrow{a}+\overrightarrow{b}$ .

+ BD = DE = EC và D, E thuộc cạnh BC nên BD = $\frac{1}{3}$ BC.

Mà $\overrightarrow{BD}$  và $\overrightarrow{BC}$  cùng hướng nên $\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BC}$ .

Suy ra: $\overrightarrow{BD}=\frac{1}{3}\left(-\overrightarrow{a}+\overrightarrow{b}\right)=-\frac{1}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$ .

Vậy $\overrightarrow{BD}=-\frac{1}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$  .

+ Hai vectơ $\overrightarrow{BE},\overrightarrow{BC}$  cùng hướng và BE = $\frac{2}{3}$ BC nên $\overrightarrow{BE}=\frac{2}{3}\overrightarrow{BC}$ .

Suy ra: $\overrightarrow{BE}=\frac{2}{3}\left(-\overrightarrow{a}+\overrightarrow{b}\right)=-\frac{2}{3}\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}$ .

Vậy $\overrightarrow{BE}=-\frac{2}{3}\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}$

+ Ta có:  $\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{a}+\left(-\frac{1}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}\right)$$=\left(1-\frac{1}{3}\right)\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$$=\frac{2}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$ .

Vậy $\overrightarrow{AD}=\frac{2}{3}\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$ .

+ Ta có: $\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{a}+\left(-\frac{2}{3}\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}\right)$$=\left(1-\frac{2}{3}\right)\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}$$=\frac{1}{3}\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}$.

Vậy $\overrightarrow{AE}=\frac{1}{3}\overrightarrow{a}+\frac{2}{3}\overrightarrow{b}$ .