NF+NB=BF
=>BF=2NB+NB=3NB
=>\(\overrightarrow{BN}=\frac13\cdot\overrightarrow{BF}\) và \(\overrightarrow{NF}=\frac23\cdot\overrightarrow{BF}\)
MC=2MA
mà MC+MA=AC
nên \(MC=\frac23AC;MA=\frac13AC\)
=>\(\overrightarrow{AM}=\frac13\cdot\overrightarrow{AC};\overrightarrow{MC}=\frac23\cdot\overrightarrow{AC}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}\)
\(=-\frac13\cdot\overrightarrow{AC}+\overrightarrow{AB}+\frac13\cdot\overrightarrow{BF}=-\frac13\cdot\overrightarrow{AC}+\overrightarrow{AB}+\frac13\left(\overrightarrow{BA}+\overrightarrow{AF}\right)\)
\(=-\frac13\cdot\overrightarrow{AC}+\overrightarrow{AB}-\frac13\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AF}=-\frac13\cdot\overrightarrow{AC}+\frac23\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AF}\)
\(=-\frac13\cdot\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+\frac23\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AF}\)
\(=\frac13\cdot\overrightarrow{AB}-\frac13\cdot\overrightarrow{AD}+\frac13\cdot\overrightarrow{AF}=\frac13\cdot\left(\overrightarrow{AB}-\overrightarrow{AD}+\overrightarrow{AF}\right)\)
\(\overrightarrow{DE}=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BE}\)
\(=\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{AF}=-\overrightarrow{AD}+\overrightarrow{AB}+\overrightarrow{AF}\)
=>\(\overrightarrow{DE}=3\cdot\overrightarrow{MN}\)
b: Vì \(\overrightarrow{DE}=3\cdot\overrightarrow{MN}\)
nên DE//MN
