a.

\(\overrightarrow{AM}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CM}+\overrightarrow{BM}+\overrightarrow{MC}=\overrightarrow{AC}+\overrightarrow{BM}\)

b.

\(\overrightarrow{AE}=3\overrightarrow{EM}=3\overrightarrow{EA}+3\overrightarrow{AM}\Rightarrow4\overrightarrow{AE}=3\overrightarrow{AM}\Rightarrow\overrightarrow{AE}=\dfrac{3}{4}\overrightarrow{AM}\)

\(\Rightarrow\overrightarrow{AE}=\dfrac{3}{4}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{3}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)

\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}=-\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}=-\dfrac{5}{8}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)

\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}=\dfrac{8}{5}\overrightarrow{BE}\)

\(\Rightarrow\) B, E, K thẳng hàng

Bài này có 1 cách cực kì nhanh, ko cần phân tích vecto, đó là sử dụng Menelaus của lớp 8:

Nếu B, K, E thẳng hàng, xét tam giác ACD có BE lần lượt cắt 3 cạnh tam giác tại E, K, B nên theo Menelaus ta có:

\(\frac{EA}{EC}.\frac{BC}{BD}.\frac{DK}{KA}=1\Leftrightarrow\frac{3}{1}.\frac{3}{1}.\frac{DK}{KA}=1\Rightarrow AK=9DK\Rightarrow AK=\frac{9}{10}AD\)

Vậy điểm K nằm ở vị trí sao cho \(\overrightarrow{AK}=\frac{9}{10}\overrightarrow{AD}\) thì B, K, E thẳng hàng

Dễ thấy: \(\overrightarrow {BC}  = \overrightarrow {BA}  + \overrightarrow {AC}  =  - \overrightarrow {AB}  + \overrightarrow {AC} \)

Ta có:

 +) \(\overrightarrow {AD}  = \overrightarrow {AB}  + \overrightarrow {BD} \). Mà \(\overrightarrow {BD}  =  - \overrightarrow {DB}  =  - \frac{1}{3}\overrightarrow {BC} \)

\( \Rightarrow \overrightarrow {AD}  = \overrightarrow {AB}  + \left( { - \frac{1}{3}} \right)( - \overrightarrow {AB}  + \overrightarrow {AC} ) = \frac{4}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {AC} \)

+) \(\overrightarrow {DH}  = \overrightarrow {DA}  + \overrightarrow {AH}  =  - \overrightarrow {AD}  + \overrightarrow {AH} \).

Mà \(\overrightarrow {AD}  = \frac{4}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {AC} ;\;\;\overrightarrow {AH}  = \frac{2}{3}\overrightarrow {AB} .\)

\( \Rightarrow \overrightarrow {DH}  =  - \left( {\frac{4}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {AC} } \right) + \frac{2}{3}\overrightarrow {AB}  =  - \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} .\)

+) \(\overrightarrow {HE}  = \overrightarrow {HA}  + \overrightarrow {AE}  =  - \overrightarrow {AH}  + \overrightarrow {AE} \)

Mà \(\overrightarrow {AH}  = \frac{2}{3}\overrightarrow {AB} ;\;\overrightarrow {AE}  = \frac{1}{3}\overrightarrow {AC} \)

\( \Rightarrow \overrightarrow {HE}  =  - \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} .\)

b)

Theo câu a, ta có: \(\overrightarrow {DH}  = \overrightarrow {HE}  =  - \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} \)

\( \Rightarrow \) Hai vecto \(\overrightarrow {DH} ,\overrightarrow {HE} \) cùng phương.

\( \Leftrightarrow \)D, E, H thẳng hàng

Ta có: \(\overrightarrow {AB}  + \overrightarrow {BC}  = \overrightarrow {AC}  \Leftrightarrow \overrightarrow {BC}  = \overrightarrow b  - \overrightarrow a \)

Lại có: vecto \(\overrightarrow {BD} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BD} } \right| = \frac{1}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BD}  = \frac{1}{3}\overrightarrow {BC}  = \frac{1}{3}(\overrightarrow b  - \overrightarrow a )\)

Tương tự: vecto \(\overrightarrow {BE} ,\overrightarrow {BC} \) cùng hướng và \(\left| {\overrightarrow {BE} } \right| = \frac{2}{3}\left| {\overrightarrow {BC} } \right|\)

\( \Rightarrow \overrightarrow {BE}  = \frac{2}{3}\overrightarrow {BC}  = \frac{2}{3}(\overrightarrow b  - \overrightarrow a )\)

Ta có:

\(\overrightarrow {AB}  + \overrightarrow {BD}  = \overrightarrow {AD}  \Leftrightarrow \overrightarrow {AD}  = \overrightarrow a  + \frac{1}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{2}{3}\overrightarrow a  + \frac{1}{3}\overrightarrow b \)

\(\overrightarrow {AB}  + \overrightarrow {BE}  = \overrightarrow {AE}  \Leftrightarrow \overrightarrow {AE}  = \overrightarrow a  + \frac{2}{3}(\overrightarrow b  - \overrightarrow a ) = \frac{1}{3}\overrightarrow a  + \frac{2}{3}\overrightarrow b \)

\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)

\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)

\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)

\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)

a)
\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IB}=\overrightarrow{AI}+\dfrac{1}{2}\left(\overrightarrow{IA}+\overrightarrow{AB}\right)\)
\(=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IA}+\dfrac{1}{2}\overrightarrow{AB}\)\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}\).
b) Theo câu a:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).