Cho tứ giác ABCD. Gọi M, N lần lượt là trung điểm của AB và CD

* Hướng dẫn giải

Ta có $\overrightarrow{MN}=\overrightarrow{MA}+\text{}\overrightarrow{AD}+\overrightarrow{DN};\overrightarrow{MN}=\overrightarrow{MB}+\text{}\overrightarrow{BC}+\overrightarrow{CN}$

$\begin{array}{l}\Rightarrow 2\overrightarrow{MN}=\overrightarrow{MA}+\text{}\overrightarrow{AD}+\overrightarrow{DN}+\text{}\overrightarrow{MB}+\text{}\overrightarrow{BC}+\overrightarrow{CN}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=(\overrightarrow{MA}+\text{}\overrightarrow{MB})+(\overrightarrow{AD}+\text{}\overrightarrow{BC})+(\text{}\overrightarrow{DN}+\overrightarrow{CN})\\ =\overrightarrow{0}+(\overrightarrow{AD}+\text{}\overrightarrow{BC})+\text{}\overrightarrow{0}=\overrightarrow{AD}+\text{}\overrightarrow{BC}\end{array}$

$\Rightarrow \overrightarrow{MN}=\frac{1}{2}\left(\overrightarrow{AD}+\text{}\overrightarrow{BC}\right)$

Đáp án C