\(\overrightarrow{IJ}=\overrightarrow{AI}+\overrightarrow{AJ}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=-\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\)
\(\overrightarrow{IJ}=-\frac{1}{6}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AB}+3\overrightarrow{IJ}\)
\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\frac{1}{2}\overrightarrow{AB}+m.\overrightarrow{IJ}\)
\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}\left(\frac{1}{2}\overrightarrow{B}+3\overrightarrow{IJ}\right)\)
\(\overrightarrow{AD}=\frac{5}{4}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{IJ}=\frac{5}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{IJ}\right)\)
Vậy để A;K;D thẳng hàng \(\Leftrightarrow m=\frac{3}{5}\)