a) ta có vector AA'+vectorBB'+vectorCC'=1/2(vectorAB+vectorAC+vectorBA+vectorBC+vectorCA+vectorCB)=vector 0

t/c trung tuyến

Xét ΔABC có AM là đường trung tuyến

nên \(\overrightarrow{AM}=\frac12\left(\overrightarrow{AB}_{}+\overrightarrow{AC}\right)\)

Xét ΔABC có BN là đường trung tuyến

nên \(\overrightarrow{BN}=\frac12\left(\overrightarrow{BC}+\overrightarrow{BA}\right)\)

Xét ΔABC có CE là đường trung tuyến

nên \(\overrightarrow{CE}=\frac12\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\)

\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CE}\)

\(=\frac12\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{BC}+\overrightarrow{BA}+\overrightarrow{CA}+\overrightarrow{CB}\right)=\overrightarrow{0}\)

\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AE}\)

\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AE}\)

\(\Rightarrow\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AN}\)

a: \(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}\)

\(=\overrightarrow{CB}+\overrightarrow{BC}\)

\(=\overrightarrow{0}\)

b: \(\overrightarrow{AM}+\overrightarrow{AP}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\cdot2\cdot\overrightarrow{AN}=\overrightarrow{AN}\)