gọi H là trực tâm tg ABC, AH giao BC tại K và AG giao BC tại M
gọi A2,B2,C2 là X,Y,Z nhé viết dưới mỏi tay quá :v
$\overrightarrow{AX}= 2\overrightarrow{AG}-\overrightarrow{AD} = 2\overrightarrow{AG}-(\overrightarrow{AG}+\overrightarrow{GD}) =\overrightarrow{AG}-\overrightarrow{GD} =\overrightarrow{AG}-\frac{1}{3}\overrightarrow{AK}$
có $\alpha \overrightarrow{HA}+\beta \overrightarrow{HB}+\gamma \overrightarrow{HC}=\overrightarrow{0} (\alpha +\beta +\gamma \neq 0) \rightarrow \beta \overrightarrow{KB}+\gamma \overrightarrow{KC}=\overrightarrow{0} \rightarrow \beta \overrightarrow{AB}+\gamma \overrightarrow{AC}=(\beta +\gamma )\overrightarrow{AK} =>\overrightarrow{AX}=1/3(\overrightarrow{AB}+\overrightarrow{AC})-1/3(\frac{\beta \overrightarrow{AB}+\gamma \overrightarrow{AC}}{\beta +\gamma }) =>3\overrightarrow{AX}=\frac{\beta \overrightarrow{AC}+\gamma \overrightarrow{AB}}{\beta +\gamma }$
dựng I thỏa mãn:$\frac{1}{\alpha }\overrightarrow{IA}+\frac{1}{\beta }\overrightarrow{IB}+\frac{1}{\gamma }\overrightarrow{IC}=\overrightarrow{0} -)\frac{1}{\beta }\overrightarrow{AB}+\frac{1}{\gamma }\overrightarrow{AC}=(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })\overrightarrow{AI} ->\gamma \overrightarrow{AB}+\beta \overrightarrow{AC}=\beta \gamma (\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })\overrightarrow{AI} -)3(\beta +\gamma )\overrightarrow{AX}=\beta \gamma (\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })\overrightarrow{AI} ->\overline{A,X,I}$
Tương tự BY,CZ đi qua I