Question

Cho tam giác ABC. Gọi D,E,K lần lượt là các điểm thoản mãn:\(\overrightarrow{BD}\)^=2/3 \(\overrightarrow{BC}\)^{, \(\overrightarrow{AE}\)=1/4 \(\overrightarrow{AC}\), \(\overrightarrow{AK}\)=1/3 \(\overrightarrow{AD}\)
Chứng minh B,K,E thẳng hàng}

Answer 1

\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AD}\)

\(=\overrightarrow{BA}+\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BD}\right)\)

\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BD}\)

\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{9}\overrightarrow{BC}\)

\(=\dfrac{8}{9}\overrightarrow{BA}+\dfrac{2}{9}\overrightarrow{AC}\)

\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}=\overrightarrow{BA}+\dfrac{1}{4}\overrightarrow{AC}\)

Vì 8/9:1=2/9:1/4

nên B,E,K thẳng hàng