Question

[Kiriya Niki]

a. Cho 6 điểm A, B, C, D, E, F. Chứng minh rằng: \(\overrightarrow{AD}\) + \(\overrightarrow{BE}\) + \(\overrightarrow{CF}\) = \(\overrightarrow{AE}\) + \(\overrightarrow{BF}\) + \(\overrightarrow{CD}\)

b. Cho AK và BM là 2 đường trung tuyến của tam giác ABC.

Hãy phân tích các vectơ \(\overrightarrow{AB}\) ; \(\overrightarrow{BC}\) ; \(\overrightarrow{AC}\) theo 2 vectơ \(\overrightarrow{u}\) = \(\overrightarrow{AK}\) ; \(\overrightarrow{v}\) = \(\overrightarrow{BM}\)

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Answer 1

[Nguyễn Việt Lâm]

a/ \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{EF}+\overrightarrow{FE}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\)

b/ Theo tính chất trung tuyến:

\(\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AK}\\\overrightarrow{BA}+\overrightarrow{BC}=2\overrightarrow{BM}\end{matrix}\right.\) \(\Rightarrow\overrightarrow{AC}+\overrightarrow{BC}=2\overrightarrow{AK}+2\overrightarrow{BM}\)

\(\overrightarrow{AC}=\overrightarrow{AK}+\overrightarrow{KC}=\overrightarrow{AK}+\frac{1}{2}\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AK}+2\overrightarrow{BM}-\frac{1}{2}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\)

\(\Rightarrow\overrightarrow{AC}=\overrightarrow{AK}+\frac{1}{2}\left(\frac{3}{2}\overrightarrow{AK}+\frac{4}{3}\overrightarrow{BM}\right)=...\)

\(\overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{BC}=...\)