Bài 9:
Vì ABCD là hình bình hành
nên \(\overrightarrow{AB}=\overrightarrow{DC}\)(1)
\(\overrightarrow{MB}-\overrightarrow{MA}=\overrightarrow{AM}+\overrightarrow{MB}=\overrightarrow{AB}\)(2)
\(\overrightarrow{MC}-\overrightarrow{MD}=\overrightarrow{DM}+\overrightarrow{MC}=\overrightarrow{DC}\)(3)
Từ (1),(2),(3) suy ra \(\overrightarrow{MB}-\overrightarrow{MA}=\overrightarrow{MC}-\overrightarrow{MD}\)
=>\(\overrightarrow{MA}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{MD}\)
Bài 10:
a:
M là trung điểm của AD nên \(\overrightarrow{MA}=-\overrightarrow{MD}\)
N là trung điểm của BC nên \(\overrightarrow{NB}=-\overrightarrow{NC}\)
=>\(\overrightarrow{BN}=-\overrightarrow{CN}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}\)
\(\overrightarrow{MN}=\overrightarrow{MD}+\overrightarrow{DC}+\overrightarrow{CN}\)
Do đó: \(2\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{MD}+\overrightarrow{DC}+\overrightarrow{CN}\)
=>\(2\overrightarrow{MN}=\left(\overrightarrow{AB}+\overrightarrow{DC}\right)+\left(\overrightarrow{MA}+\overrightarrow{MD}\right)+\left(\overrightarrow{BN}+\overrightarrow{CN}\right)\)
\(\Leftrightarrow2\cdot\overrightarrow{MN}=\overrightarrow{AB}+\overrightarrow{DC}\)
b: \(\overrightarrow{AC}+\overrightarrow{DB}\)
\(=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{CB}\)
\(=\overrightarrow{AB}+\overrightarrow{DC}=2\overrightarrow{MN}\)
