Câu hỏi:

Gọi E, F lần lượt là trung điểm của AB, CD.
O là trung điểm của EF.
Ta có:
\(\overrightarrow{OA} + \overrightarrow{OB} = 2\overrightarrow{OE}\) (1)
\(\overrightarrow{OC} + \overrightarrow{OD} = 2\overrightarrow{OF}\) (2)
Từ (1) và (2) suy ra:
\(\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 2\overrightarrow{OE} + 2\overrightarrow{OF} = 2(\overrightarrow{OE} + \overrightarrow{OF}) = 2.\overrightarrow{0} = \overrightarrow{0}\)
Vậy \(\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} = \overrightarrow 0 \)

Ta có: $AB = 3AM \Rightarrow \overrightarrow{AM} = \frac{1}{3} \overrightarrow{AB} = \frac{1}{3} \overrightarrow{a}$

$CN = 2BC \Rightarrow \overrightarrow{CN} = 2\overrightarrow{BC} = 2(\overrightarrow{AC} - \overrightarrow{AB}) = 2(\overrightarrow{b} - \overrightarrow{a}) \Rightarrow \overrightarrow{AN} = \overrightarrow{AC} + \overrightarrow{CN} = \overrightarrow{b} + 2(\overrightarrow{b} - \overrightarrow{a}) = 3\overrightarrow{b} - 2\overrightarrow{a} = -2\overrightarrow{a} + 3\overrightarrow{b}$

Ta có: $\overrightarrow{y} = 2\overrightarrow{a} + 6\overrightarrow{b} = 2(\overrightarrow{a} + 3\overrightarrow{b}) = -2(\overrightarrow{a} - (-3)\overrightarrow{b}) = -2(\overrightarrow{a} - 3\overrightarrow{b}) = -2\overrightarrow{x}$.
Vì $\overrightarrow{y} = -2\overrightarrow{x}$ nên $\overrightarrow{y}$ và $\overrightarrow{x}$ cùng phương, ngược hướng.

Ta có $\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB} = \frac{3}{4}(\overrightarrow{BC} - \overrightarrow{BA}) - \overrightarrow{AB} = \frac{3}{4}\overrightarrow{BC} - \frac{3}{4}\overrightarrow{BA} - \overrightarrow{AB} = \frac{3}{4}\overrightarrow{BC} - \frac{1}{4}\overrightarrow{AB}$.

Suy ra $\overrightarrow{BI} = \frac{3}{4}\overrightarrow{BC} + \frac{1}{4}\overrightarrow{BA}$.

$\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{BC} - \overrightarrow{BA}) - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{BC} - \frac{1}{2}\overrightarrow{BA} - \frac{2}{3}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{BC} - \frac{1}{6}\overrightarrow{AB}$.

Suy ra $\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{BC} + \frac{1}{6}\overrightarrow{BA}$.

$\overrightarrow{BJ} = x \overrightarrow{BI} \Leftrightarrow \frac{1}{2}\overrightarrow{BC} + \frac{1}{6}\overrightarrow{BA} = x(\frac{3}{4}\overrightarrow{BC} + \frac{1}{4}\overrightarrow{BA}) \Leftrightarrow \frac{1}{2} = \frac{3}{4}x \land \frac{1}{6} = \frac{1}{4}x \Leftrightarrow x = \frac{2}{3}$.

Do đó $\overrightarrow{BJ} = \frac{2}{3}\overrightarrow{BI}$.

Vậy B, I, J thẳng hàng.

Tuy nhiên, đáp án này không đúng. Kiểm tra lại đề bài.

$\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB}$

$\overrightarrow{BJ} = \frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB}$

$\overrightarrow{IJ} = \overrightarrow{BJ} - \overrightarrow{BI} = (\frac{1}{2}\overrightarrow{AC} - \frac{2}{3}\overrightarrow{AB}) - (\frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB}) = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB}$

$\overrightarrow{IC} = \overrightarrow{AC} - \overrightarrow{AI}$. Ta cần tìm hệ số $k$ sao cho $\overrightarrow{AI} = k \overrightarrow{AC}$

$\overrightarrow{BI} = \frac{3}{4}\overrightarrow{AC} - \overrightarrow{AB} = \overrightarrow{AI} - \overrightarrow{AB} = k\overrightarrow{AC} - \overrightarrow{AB}$

$\Rightarrow k = \frac{3}{4} \Rightarrow \overrightarrow{AI} = \frac{3}{4}\overrightarrow{AC}$

$\overrightarrow{IC} = \overrightarrow{AC} - \frac{3}{4}\overrightarrow{AC} = \frac{1}{4}\overrightarrow{AC}$

$\overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -\overrightarrow{IC} + \frac{1}{3}\overrightarrow{AB}$

$\Rightarrow \overrightarrow{CJ} = \overrightarrow{CI} + \overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} - \frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -\frac{1}{2}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = -(\frac{1}{2}\overrightarrow{AC} - \frac{1}{3}\overrightarrow{AB}) = -\overrightarrow{BJ}$

Vậy J là trung điểm BC.

$\overrightarrow{IJ} = -\frac{1}{4}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AB} = k \overrightarrow{IC} = k \frac{1}{4}\overrightarrow{AC}$

$\overrightarrow{IJ} = x \overrightarrow{JC}$

Vậy I, J, C thẳng hàng.