Cho tam giác ABC, có ba trung tuyến AD, BE, CF. Chứng minh rằng

Giải SBT Toán 10 Cánh diều Bài ôn tập chương 4

Bài 80 trang 108 SBT Toán 10 Tập 1: Cho tam giác ABC, có ba trung tuyến AD, BE, CF. Chứng minh rằng: $\overrightarrow{AD}.\overrightarrow{BC}+\overrightarrow{BE}.\overrightarrow{CA}+\overrightarrow{CF}.\overrightarrow{AB}=0$.

Lời giải:

Ta có: $\overrightarrow{AD}.\overrightarrow{BC}+\overrightarrow{BE}.\overrightarrow{CA}+\overrightarrow{CF}.\overrightarrow{AB}$

= $\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right).\overrightarrow{BC}+\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BC}\right).$$\overrightarrow{CA}+\frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right).\overrightarrow{AB}$

= $\frac{1}{2}\overrightarrow{AB}.\overrightarrow{BC}+\frac{1}{2}\overrightarrow{AC}.\overrightarrow{BC}+\frac{1}{2}\overrightarrow{BA}.\overrightarrow{CA}$$+\frac{1}{2}\overrightarrow{BC}.\overrightarrow{CA}+\frac{1}{2}\overrightarrow{CA}.\overrightarrow{AB}+\frac{1}{2}\overrightarrow{CB}.\overrightarrow{AB}$

= $\left(\frac{1}{2}\overrightarrow{AB}.\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CB}.\overrightarrow{AB}\right)+$$\left(\frac{1}{2}\overrightarrow{AC}.\overrightarrow{BC}+\frac{1}{2}\overrightarrow{BC}.\overrightarrow{CA}\right)+$$\left(\frac{1}{2}\overrightarrow{BA}.\overrightarrow{CA}+\frac{1}{2}\overrightarrow{CA}.\overrightarrow{AB}\right)$

= 0