Kết quả tìm kiếm

\frac{DB}{BA} = \frac{DC}{AC} = \frac{DC+DB}{BA+AC} = \frac{a}{b+c} \\ \\ DB = BA.\frac{a}{b+c} \Rightarrow DB = \frac{a.c}{b+c} \\ \\ \Rightarrow \vec{BD} = \frac{c}{b+c}\vec{BC} \\ \\ ta-co: \vec{AD} = \vec{AB} +\vec{BD} = \vec{AB} + \frac{c}{b+c}\vec{BC} \\ \\ = \vec{AB} -...