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\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=-\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}=-\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=-\overrightarrow{AB}+\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)\)
\(\overrightarrow{BM}=-\frac{3}{4}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}=\frac{3}{4}\left(-\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)=\frac{3}{4}\overrightarrow{BN}\)
\(\Rightarrow B;M;N\) thẳng hàng
Xét ΔBAD có BI là đường trung tuyến
nên \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)
=>\(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1}{3}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)
\(\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}\)
\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\)
=>\(\overrightarrow{BI}=\dfrac{5}{6}\cdot\overrightarrow{BM}\)
=>B,I,M thẳng hàng
Cách 1: Dùng định lý Menelaus đảo:
Từ đề bài, ta có \(\dfrac{BD}{BC}=\dfrac{2}{3}\), \(\dfrac{MC}{MA}=\dfrac{3}{2}\), \(\dfrac{IA}{ID}=1\)
\(\Rightarrow\dfrac{BD}{BC}.\dfrac{MC}{MA}.\dfrac{IA}{ID}=1\)
Theo định lý Menelaus đảo, suy ra B, I, M thẳng hàng.
Cách 2: Dùng vector
Ta có \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{2}{3}\overrightarrow{BC}\)
\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)
Lại có \(\overrightarrow{BM}=\dfrac{MC}{AC}\overrightarrow{BA}+\dfrac{MA}{AC}\overrightarrow{BC}\)
\(=\dfrac{3}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)
\(=\dfrac{1}{5}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)
\(=\dfrac{6}{5}.\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)
\(=\dfrac{6}{5}\overrightarrow{BI}\)
Vậy \(\overrightarrow{BM}=\dfrac{6}{5}\overrightarrow{BI}\), suy ra B, I, M thẳng hàng.
a: \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}-\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
Xét ΔBAD có BM là đường trung tuyến
nên \(\overrightarrow{BM}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)\)
\(=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)
\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}\)
\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)
=>\(\overrightarrow{BM}=\dfrac{5}{6}\cdot\overrightarrow{BN}\)
=>B,M,N thẳng hàng