\(\overrightarrow{AB}+\overrightarrow{AF}+\overrightarrow{AO}=\overrightarrow{AO}+\overrightarrow{AO}=2\overrightarrow{AO}=\overrightarrow{AD}\)
b/ Do lục giác đều nên: \(\left\{{}\begin{matrix}\overrightarrow{DC}=\overrightarrow{OB}\\\overrightarrow{FE}=\overrightarrow{AO}\end{matrix}\right.\)
\(\overrightarrow{MA}+\overrightarrow{MC}+\overrightarrow{ME}=\overrightarrow{MB}+\overrightarrow{BA}+\overrightarrow{MD}+\overrightarrow{DC}+\overrightarrow{MF}+\overrightarrow{FE}\)
\(=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{MF}+\overrightarrow{BA}+\overrightarrow{AO}+\overrightarrow{OB}\)
\(=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{MF}+\overrightarrow{BO}+\overrightarrow{OB}=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{MF}\)