Question

Cho tam giác ABC, M là trung điểm AB. N ϵ AC, NC=2NA, K là trung điểm MN, D là trung điểm BC. Chứng minh vectoKD = 1/4 vectoAB + 1/3vectoAC

Answer 1

Có \(NC=2NA\Rightarrow\overrightarrow{NA}=\frac{1}{3}\overrightarrow{CA}\)

Có \(\overrightarrow{KD}=\overrightarrow{KM}+\overrightarrow{MB}+\overrightarrow{BD}\)

\(=\frac{1}{2}\overrightarrow{NM}+\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}\)

\(=\frac{1}{2}\left(\overrightarrow{NA}+\overrightarrow{AM}\right)+\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{2}\overrightarrow{AC}\)

\(=\frac{1}{2}\left(\frac{1}{3}\overrightarrow{CA}+\frac{1}{2}\overrightarrow{AB}\right)+\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{2}\overrightarrow{AC}\)

\(=\frac{1}{3}\overrightarrow{AC}+\frac{1}{4}\overrightarrow{AB}\)