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Nối AC, trên cạnh AC lấy điểm I sao cho \(\overrightarrow{AI}=\frac{2}{3}\overrightarrow{AC}\)
Xét tam giác ABC có: \(\frac{AM}{AB}=\frac{AI}{AC}=\frac{2}{3}\) \(\Rightarrow\overrightarrow{MI}=\frac{2}{3}\overrightarrow{BC}\)
Tương tự trong tam giác ACD có: \(\overrightarrow{IN}=\frac{2}{3}\overrightarrow{AD}\)
Ta có: \(\overrightarrow{MN}=\overrightarrow{MI}+\overrightarrow{IN}=\frac{2}{3}\left(\overrightarrow{BC}+\overrightarrow{AD}\right)\)
a) ta có : \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NB}+\overrightarrow{DM}+\overrightarrow{MN}+\overrightarrow{NC}\)
\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{DM}\right)+\left(\overrightarrow{NB}+\overrightarrow{NC}\right)=2\overrightarrow{MN}\left(đpcm\right)\)
b) ta có : \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AI}+\overrightarrow{IJ}+\overrightarrow{JB}+\overrightarrow{CI}+\overrightarrow{IJ}+\overrightarrow{JD}\)
\(=2\overrightarrow{IJ}+\left(\overrightarrow{AI}+\overrightarrow{CI}\right)+\left(\overrightarrow{JB}+\overrightarrow{JD}\right)=2\overrightarrow{IJ}\left(đpcm\right)\)
bn dùng định lí ta lét chứng minh được \(\overrightarrow{MJ}=\overrightarrow{IN}=\dfrac{1}{2}\overrightarrow{AB}\)
C) ta có : \(\overrightarrow{MN}+\overrightarrow{IJ}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{IA}+\overrightarrow{AB}+\overrightarrow{BJ}\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{MA}+\overrightarrow{BJ}\right)+\left(\overrightarrow{BN}+\overrightarrow{IA}\right)\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{DM}+\overrightarrow{JD}\right)+\left(\overrightarrow{NC}+\overrightarrow{CI}\right)=2\overrightarrow{AB}+\overrightarrow{JM}+\overrightarrow{NI}\) \(=2\overrightarrow{AB}+\overrightarrow{BA}=\overrightarrow{AB}\left(đpcm\right)\)d) ta có : \(\overrightarrow{IM}+\overrightarrow{IN}=\overrightarrow{IJ}+\overrightarrow{JM}+\overrightarrow{IN}=\overrightarrow{IJ}\left(đpcm\right)\)
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\Leftrightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\Rightarrow\overrightarrow{KA}=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{KA}+\overrightarrow{AD}=\left(-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\right)+\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow m-n=-\frac{1}{12}\)
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\)
\(\Leftrightarrow-\overrightarrow{AC}=-3\overrightarrow{AN}\Rightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}\) (do M là trung điểm AB)
\(\overrightarrow{AK}=\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{6}\end{matrix}\right.\)
a) Chữa đề: \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(Ta\text{ }có:\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{DA}+\overrightarrow{AB}\\ =\overrightarrow{CB}+\overrightarrow{DA}+\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\)\(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{DC}\\ =2\overrightarrow{CM}+2\overrightarrow{NC}=2\left(\overrightarrow{NC}+\overrightarrow{CM}\right)=2\overrightarrow{NM}\)
Vậy \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(\text{b) }\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{BC}=-\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{CA}+\overrightarrow{CB}\right)\\ =-\left[\left(\overrightarrow{DA}+\overrightarrow{DB}\right)+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\right]\\ =-\left(2\overrightarrow{DM}+2\overrightarrow{CM}\right)=2\left(\overrightarrow{MD}+\overrightarrow{MC}\right)=4\left(\overrightarrow{MN}\right)\)
\(\text{c) }2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{DA}\right)+\left(\overrightarrow{AI}+\overrightarrow{NA}\right)\right]\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{DB}\right)+\overrightarrow{NI}\right]=2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)\)
Mà IN là dường trung bình \(\Delta BCD\)
\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)