Ta có: \(\overrightarrow{CM}=2\cdot\overrightarrow{MB}\)
=>CM=2MB và M nằm giữa C và B
CM+MB=CB
=>CB=2MB+MB=3MB
=>\(\overrightarrow{BC}=3\cdot\overrightarrow{BM}\)
Ta có: CM=2MB; CB=3MB
Do đó: \(CM=\frac23CB\)
\(\overrightarrow{MN}+2\cdot\overrightarrow{DN}=\overrightarrow{0}\)
=>\(\overrightarrow{MN}=-2\cdot\overrightarrow{DN}\)
=>\(\overrightarrow{MN}=2\cdot\overrightarrow{ND}\)
=>MN=2ND và N nằm giữa M và D
MD=MN+ND=2ND+ND=3ND
=>\(\overrightarrow{DM}=3\cdot\overrightarrow{DN}\)
\(\overrightarrow{CK}=\frac47\cdot\overrightarrow{CD}\)
=>\(CK=\frac47CD\) và K nằm giữa C và D
CK+KD=CD
=>\(KD=CD-CK=CD-\frac47CD=\frac37CD\)
\(\overrightarrow{AN}=\overrightarrow{AD}+\overrightarrow{DN}\)
\(=\overrightarrow{AD}+\frac13\cdot\overrightarrow{DM}=\overrightarrow{AD}+\frac13\left(\overrightarrow{DC}+\overrightarrow{CM}\right)\)
\(=\overrightarrow{AD}+\frac13\left(\overrightarrow{DC}+\frac23\cdot\overrightarrow{CB}\right)=\overrightarrow{AD}+\frac13\left(\overrightarrow{DC}-\frac23\cdot\overrightarrow{AD}\right)\)
\(=\frac13\cdot\overrightarrow{DC}+\overrightarrow{AD}-\frac29\cdot\overrightarrow{AD}=\frac13\cdot\overrightarrow{DC}+\frac79\cdot\overrightarrow{AD}\)
\(=\frac79\left(\overrightarrow{AD}+\frac37\cdot\overrightarrow{DC}\right)\)
\(\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{DK}=\overrightarrow{AD}+\frac37\cdot\overrightarrow{DC}\)
=>\(\overrightarrow{AN}=\frac79\cdot\overrightarrow{AK}\)
=>A,N,K thẳng hàng
