a) Tính \(\overset{\rightarrow}{I J} ; \overset{\rightarrow}{I K}\) theo \(\overset{\rightarrow}{A B} ; \overset{\rightarrow}{A C}\).
Ta có: \(\overset{\rightarrow}{I J} = \overset{\rightarrow}{I C} + \overset{\rightarrow}{C J} = - \overset{\rightarrow}{B C} - \frac{1}{3} \overset{\rightarrow}{A C} = - \left(\right. \overset{\rightarrow}{B A} + \overset{\rightarrow}{A C} \left.\right) - \frac{1}{3} \overset{\rightarrow}{A C} = \overset{\rightarrow}{A B} - \frac{4}{3} \overset{\rightarrow}{A C}\).
\(\overset{\rightarrow}{I K} = \overset{\rightarrow}{I B} + \overset{\rightarrow}{B K} = - 2 \overset{\rightarrow}{B C} - \frac{1}{2} \overset{\rightarrow}{A B} = - 2 \left(\right. \overset{\rightarrow}{B A} + \overset{\rightarrow}{A C} \left.\right) - \frac{1}{2} \overset{\rightarrow}{A B} = \frac{3}{2} \overset{\rightarrow}{A B} - 2 \overset{\rightarrow}{A C} .\)
b) Chứng minh ba điểm \(I , J , K\) thẳng hàng.
Theo câu a: \({\overset{\rightarrow}{I J}=\overset{\rightarrow}{A B}-\frac{4}{3}\overset{\rightarrow}{A C}\\\overset{\rightarrow}{I K}=\frac{3}{2}\overset{\rightarrow}{A B}-2\overset{\rightarrow}{A C}\Leftrightarrow{\overset{\rightarrow}{I J}=\overset{\rightarrow}{A B}-\frac{4}{3}\overset{\rightarrow}{A C}\\\overset{\rightarrow}{I K}=\frac{3}{2}\left(\right.\overset{\rightarrow}{A B}-\frac{2}{4}\overset{\rightarrow}{A C}\left.\right)\Rightarrow\overset{\rightarrow}{I K}=\frac{3}{2}\overset{\rightarrow}{I J}.}}\) \(\Rightarrow I , J , K\) thẳng hàng.

Ta có \(\overset{\rightarrow}{B C} + \overset{\rightarrow}{M A} = \overset{\rightarrow}{0} \Leftrightarrow \overset{\rightarrow}{M A} = - \overset{\rightarrow}{B C}\) nên \(M A / / B C\).
Do đó \(M \notin A C\) (1).
Ta có \(\overset{\rightarrow}{A B} - \overset{\rightarrow}{N A} - 3 \overset{\rightarrow}{A C} = \overset{\rightarrow}{0}\)
\(\Leftrightarrow \overset{\rightarrow}{A B} - \left(\right. \overset{\rightarrow}{N M} + \overset{\rightarrow}{M A} \left.\right) - 3 \overset{\rightarrow}{A C} = \overset{\rightarrow}{0}\)
\(\Leftrightarrow \overset{\rightarrow}{A B} - \overset{\rightarrow}{N M} - \overset{\rightarrow}{M A} - 3 \overset{\rightarrow}{A C} = \overset{\rightarrow}{0}\)
\(\Leftrightarrow \overset{\rightarrow}{N M} = \overset{\rightarrow}{A B} - \overset{\rightarrow}{M A} - 3 \overset{\rightarrow}{A C}\)
\(\Leftrightarrow \overset{\rightarrow}{N M} = \overset{\rightarrow}{A B} + \overset{\rightarrow}{B C} - 3 \overset{\rightarrow}{A C} = \overset{\rightarrow}{A C} - 3 \overset{\rightarrow}{A C} = - 2 \overset{\rightarrow}{A C}\) (2).
Từ \(\left(\right. 1 \left.\right) , \left(\right. 2 \left.\right)\) ta có \(M N / / A C\).

 Ta có \(\overset{\rightarrow}{B I} = \frac{1}{2} \left(\right. \overset{\rightarrow}{B A} + \overset{\rightarrow}{B M} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{B A} + \frac{1}{2} \overset{\rightarrow}{B C} \left.\right)\)

\(= \frac{1}{2} \overset{\rightarrow}{B A} + \frac{1}{4} \overset{\rightarrow}{B C} = \frac{1}{2} \left(\right. \overset{\rightarrow}{B K} + \overset{\rightarrow}{K A} \left.\right) + \frac{1}{4} \left(\right. \overset{\rightarrow}{B K} + \overset{\rightarrow}{K C} \left.\right) = \frac{3}{4} \overset{\rightarrow}{B K} + \frac{1}{2} \overset{\rightarrow}{K A} + \frac{1}{4} \overset{\rightarrow}{K C}\)
Mà \(A K = \frac{1}{3} A C\) nên \(K C = 2 K A\) suy ra \(\overset{\rightarrow}{K C} = - 2 \overset{\rightarrow}{K A} \Leftrightarrow \overset{\rightarrow}{K C} + 2 \overset{\rightarrow}{K A} = \overset{\rightarrow}{0} \Leftrightarrow \frac{1}{4} \overset{\rightarrow}{K C} + \frac{1}{2} \overset{\rightarrow}{K A} = \overset{\rightarrow}{0} .\)
Do đó \(\overset{\rightarrow}{B I} = \frac{3}{4} \overset{\rightarrow}{B K} + \overset{\rightarrow}{0} = \frac{3}{4} \overset{\rightarrow}{B K}\). Vậy ba điểm \(B , I , K\) thẳng hàng.

a) Chứng minh rằng \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right)\).

- Chứng minh \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right)\).

Vì \(M\) là trung điểm của \(A D\) nên \(\overset{\rightarrow}{M A} + \overset{\rightarrow}{M D} = \overset{\rightarrow}{0}\).

Vì \(N\) là trung điểm của \(B C\) nên \(\overset{\rightarrow}{B N} + \overset{\rightarrow}{C N} = \overset{\rightarrow}{0}\).
Áp dụng quy tắc ba điểm, ta có:

\({\overset{\rightarrow}{M N}=\overset{\rightarrow}{M A}+\overset{\rightarrow}{A B}+\overset{\rightarrow}{B N}\\\overset{\rightarrow}{M N}=\overset{\rightarrow}{M D}+\overset{\rightarrow}{D C}+\overset{\rightarrow}{C N}}\) \(\Rightarrow 2 \overset{\rightarrow}{M N} = \left(\right. \overset{\rightarrow}{M A} + \overset{\rightarrow}{M D} \left.\right) + \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} + \left(\right. \overset{\rightarrow}{B N} + \overset{\rightarrow}{C N} \left.\right) = \overset{\rightarrow}{0} + \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} + \overset{\rightarrow}{0} = \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} .\) \(\Rightarrow \overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) .\)

- Chứng minh \(\frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right) .\) \({\overset{\rightarrow}{A B}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{C B}=\overset{\rightarrow}{D B}=\overset{\rightarrow}{C D}+\overset{\rightarrow}{B C}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{D B}+\overset{\rightarrow}{C B}+\overset{\rightarrow}{B C}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{D B}\Rightarrow.}\)

Vậy: \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right)\).

b) Gọi \(I\) là trung điểm của \(M N\). Chứng minh rằng: \(\overset{\rightarrow}{I A} + \overset{\rightarrow}{I B} + \overset{\rightarrow}{I C} + \overset{\rightarrow}{I D} = \overset{\rightarrow}{0}\).

Áp dụng hệ thức trung điểm, ta có:

\({\overset{\rightarrow}{I A}+\overset{\rightarrow}{I D}=2\overset{\rightarrow}{I M}\\\overset{\rightarrow}{I B}+\overset{\rightarrow}{I D}=2\overset{\rightarrow}{I N}\Rightarrow\overset{\rightarrow}{I A}+\overset{\rightarrow}{I D}+\overset{\rightarrow}{I B}+\overset{\rightarrow}{I D}=2\left(\right.\overset{\rightarrow}{I M}+\overset{\rightarrow}{I N}\left.\right)=2.\overset{\rightarrow}{0}=\overset{\rightarrow}{0}.}\)

a) Chứng minh rằng \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right)\).

- Chứng minh \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right)\).

Vì \(M\) là trung điểm của \(A D\) nên \(\overset{\rightarrow}{M A} + \overset{\rightarrow}{M D} = \overset{\rightarrow}{0}\).

Vì \(N\) là trung điểm của \(B C\) nên \(\overset{\rightarrow}{B N} + \overset{\rightarrow}{C N} = \overset{\rightarrow}{0}\).
Áp dụng quy tắc ba điểm, ta có:

\({\overset{\rightarrow}{M N}=\overset{\rightarrow}{M A}+\overset{\rightarrow}{A B}+\overset{\rightarrow}{B N}\\\overset{\rightarrow}{M N}=\overset{\rightarrow}{M D}+\overset{\rightarrow}{D C}+\overset{\rightarrow}{C N}}\) \(\Rightarrow 2 \overset{\rightarrow}{M N} = \left(\right. \overset{\rightarrow}{M A} + \overset{\rightarrow}{M D} \left.\right) + \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} + \left(\right. \overset{\rightarrow}{B N} + \overset{\rightarrow}{C N} \left.\right) = \overset{\rightarrow}{0} + \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} + \overset{\rightarrow}{0} = \overset{\rightarrow}{A B} + \overset{\rightarrow}{C D} .\) \(\Rightarrow \overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) .\)

- Chứng minh \(\frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right) .\) \({\overset{\rightarrow}{A B}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{C B}=\overset{\rightarrow}{D B}=\overset{\rightarrow}{C D}+\overset{\rightarrow}{B C}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{D B}+\overset{\rightarrow}{C B}+\overset{\rightarrow}{B C}=\overset{\rightarrow}{A C}+\overset{\rightarrow}{D B}\Rightarrow.}\)

Vậy: \(\overset{\rightarrow}{M N} = \frac{1}{2} \left(\right. \overset{\rightarrow}{A B} + \overset{\rightarrow}{D C} \left.\right) = \frac{1}{2} \left(\right. \overset{\rightarrow}{A C} + \overset{\rightarrow}{D B} \left.\right)\).

b) Gọi \(I\) là trung điểm của \(M N\). Chứng minh rằng: \(\overset{\rightarrow}{I A} + \overset{\rightarrow}{I B} + \overset{\rightarrow}{I C} + \overset{\rightarrow}{I D} = \overset{\rightarrow}{0}\).

Áp dụng hệ thức trung điểm, ta có:

\({\overset{\rightarrow}{I A}+\overset{\rightarrow}{I D}=2\overset{\rightarrow}{I M}\\\overset{\rightarrow}{I B}+\overset{\rightarrow}{I D}=2\overset{\rightarrow}{I N}\Rightarrow\overset{\rightarrow}{I A}+\overset{\rightarrow}{I D}+\overset{\rightarrow}{I B}+\overset{\rightarrow}{I D}=2\left(\right.\overset{\rightarrow}{I M}+\overset{\rightarrow}{I N}\left.\right)=2.\overset{\rightarrow}{0}=\overset{\rightarrow}{0}.}\)

a) \(\overset{\rightarrow}{A C} + \overset{\rightarrow}{B D} = \overset{\rightarrow}{A D} + \overset{\rightarrow}{B C} = 2 \overset{\rightarrow}{E F}\)

- \(\overset{\rightarrow}{A C} + \overset{\rightarrow}{B D} = 2 \overset{\rightarrow}{E F} \left(\right. 1 \left.\right)\).
Do \(E\) là trung điểm \(A B\) nên \(2 \overset{\rightarrow}{O E} = \overset{\rightarrow}{O A} + \overset{\rightarrow}{O B}\) với \(O\) là một điểm tùy ý.

Do \(F\) là trung điểm \(C D\) nên \(2 \overset{\rightarrow}{O F} = \overset{\rightarrow}{O C} + \overset{\rightarrow}{O D}\) với \(O\) là một điểm tùy ý.

(1) \(\Leftrightarrow \overset{\rightarrow}{O C} - \overset{\rightarrow}{O A} + \overset{\rightarrow}{O D} - \overset{\rightarrow}{O B} = 2 \overset{\rightarrow}{O F} - 2 \overset{\rightarrow}{O E}\)
\(\&\Leftrightarrow\overset{\rightarrow}{O C}-\overset{\rightarrow}{O A}+\overset{\rightarrow}{O D}-\overset{\rightarrow}{O B}=\left(\right.\overset{\rightarrow}{O C}+\overset{\rightarrow}{O D}\left.\right)-\left(\right.\overset{\rightarrow}{O A}+\overset{\rightarrow}{O B}\left.\right)\\\&\Leftrightarrow\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O C} - \overset{\rightarrow}{O C}}}\left.\right)+\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O D} - \overset{\rightarrow}{O D}}}\left.\right)-\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O B} - \overset{\rightarrow}{O B}}}\left.\right)+\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O A} - \overset{\rightarrow}{O A}}}\left.\right)=\overset{\rightarrow}{0}\Rightarrow\)
\(\overset{\rightarrow}{A D} + \overset{\rightarrow}{B C} = 2 \overset{\rightarrow}{E F} \left(\right. 2 \left.\right)\)

Do \(E\) là trung điểm \(A B\) nên \(2 \overset{\rightarrow}{O E} = \overset{\rightarrow}{O A} + \overset{\rightarrow}{O B}\) với \(O\) là một điểm tùy ý.

Do \(F\) là trung điểm \(C D\) nên \(2 \overset{\rightarrow}{O F} = \overset{\rightarrow}{O C} + \overset{\rightarrow}{O D}\) với \(O\) là một điểm tùy ý.

\(\left(\right. 2 \left.\right) \Leftrightarrow \overset{\rightarrow}{O D} - \overset{\rightarrow}{O A} + \overset{\rightarrow}{O C} - \overset{\rightarrow}{O B} = 2 \overset{\rightarrow}{O F} - 2 \overset{\rightarrow}{O E}\)
\(\Leftrightarrow \overset{\rightarrow}{O D} - \overset{\rightarrow}{O A} + \overset{\rightarrow}{O C} - \overset{\rightarrow}{O B} = \left(\right. \overset{\rightarrow}{O C} + \overset{\rightarrow}{O D} \left.\right) - \left(\right. \overset{\rightarrow}{O A} + \overset{\rightarrow}{O B} \left.\right)\)
\(\Leftrightarrow\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O C} - \overset{\rightarrow}{O C}}}\left.\right)+\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O D} - \overset{\rightarrow}{O D}}}\left.\right)-\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O B} - \overset{\rightarrow}{O B}}}\left.\right)+\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{O A} - \overset{\rightarrow}{O A}}}\left.\right)=\overset{\rightarrow}{0}\)
b) \(\overset{\rightarrow}{G A} + \overset{\rightarrow}{G B} + \overset{\rightarrow}{G C} + \overset{\rightarrow}{G D} = \overset{\rightarrow}{0}\) (3).

Do \(E\) là trung điểm \(A B\) nên \(2 \overset{\rightarrow}{O E} = \overset{\rightarrow}{O A} + \overset{\rightarrow}{O B}\) với \(O\) là một điểm tùy ý.

Do \(F\) là trung điểm \(C D\) nên \(2 \overset{\rightarrow}{O F} = \overset{\rightarrow}{O C} + \overset{\rightarrow}{O D}\) với \(O\) là một điểm tùy ý.

\(\left(\right. 3 \left.\right) \Leftrightarrow \left(\right. 2 \overset{\rightarrow}{G E} - \overset{\rightarrow}{G B} \left.\right) + \overset{\rightarrow}{G B} + \overset{\rightarrow}{G C} + \left(\right. 2 \overset{\rightarrow}{G F} - \overset{\rightarrow}{G C} \left.\right) = \overset{\rightarrow}{0}\)
\(\Leftrightarrow2\overset{\rightarrow}{G E}+2\overset{\rightarrow}{G F}=\overset{\rightarrow}{0}\Leftrightarrow2\left(\right.\underset{\overset{\rightarrow}{0}}{\underbrace{\overset{\rightarrow}{G E} + \overset{\rightarrow}{G F}}}\left.\right)=\overset{\rightarrow}{0}\)

\(\overrightarrow{AB}\) = \(\overrightarrow{b}\) - \(\overrightarrow{a}\)

\(\overrightarrow{BC}\) =\(\overrightarrow{-a}\) \(-\overrightarrow{2b}\)