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\(\overrightarrow{MA}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{BA}+\overrightarrow{MD}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}\)
b/
\(2\left(\overrightarrow{JA}+\overrightarrow{AB}+\overrightarrow{DA}+\overrightarrow{AI}\right)=2\left(\overrightarrow{JB}+\overrightarrow{DI}\right)=2\left(\overrightarrow{JD}+\overrightarrow{DB}+\overrightarrow{DB}+\overrightarrow{BI}\right)\)
\(=2\left(2\overrightarrow{DB}+\overrightarrow{IC}+\overrightarrow{CJ}\right)=2\left(2\overrightarrow{DB}+\overrightarrow{IJ}\right)=2\left(2\overrightarrow{DB}+\frac{1}{2}\overrightarrow{BD}\right)=3\overrightarrow{DB}\)c/
\(\overrightarrow{AK}=\overrightarrow{AB}+\overrightarrow{BK}=\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BD}=\overrightarrow{AB}+\frac{1}{6}\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\)
\(\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}+\frac{1}{5}\overrightarrow{BC}=\frac{6}{5}\left(\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\right)=\frac{6}{5}\overrightarrow{AK}\)
\(\Rightarrow A;K;H\) thẳng hàng