Phần a "tính" thì không chuẩn lắm, nghĩa là biểu diễn tổng vector theo các cạnh hình bình hành hả bạn?
\(\bullet \overrightarrow{NC}+\overrightarrow{MC}=\overrightarrow{NM}+\overrightarrow{MC}+\overrightarrow{MC}\)
\(=\overrightarrow{NM}+2\overrightarrow{MC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\)
\(\bullet \overrightarrow{AM}+\overrightarrow{CD}=\overrightarrow{AM}+\overrightarrow{BA}\)
\(=\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{BA}\)
\(=\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BC}\)
\(\bullet\overrightarrow{AD}+\overrightarrow{CN}=2\overrightarrow{ND}+\overrightarrow{CD}+\overrightarrow{DN}\)
\(=\overrightarrow{ND}+\overrightarrow{CD}=\frac{1}{2}\overrightarrow{AD}+\overrightarrow{CD}\)
b)
Ta có:
\(\overrightarrow{AM}+\overrightarrow{AN}=(\overrightarrow{AB}+\overrightarrow{BM})+(\overrightarrow{AD}+\overrightarrow{DN})\)
\(=\overrightarrow{AB}+\overrightarrow{AD}+\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{DA})\)
Do $ABCD$ là hình bình hành nên \(\overrightarrow{BC}=\overrightarrow{AD}=-\overrightarrow{DA}\)
\(\Rightarrow \overrightarrow {BC}+\overrightarrow{DA}=\overrightarrow{0}\)
Suy ra \(\Rightarrow \overrightarrow{AM}+\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{AD}\) (đpcm)
