a: M là trung điểm của AD
=>\(\overrightarrow{AM}=\overrightarrow{MD}\)
=>\(-\overrightarrow{MA}=\overrightarrow{MD}\)
=>\(\overrightarrow{MA}+\overrightarrow{MD}=\overrightarrow{0}\)
N là trung điểm của BC
=>\(\overrightarrow{BN}=\overrightarrow{NC}\)
=>\(\overrightarrow{BN}=-\overrightarrow{CN}\)
=>\(\overrightarrow{BN}+\overrightarrow{CN}=\overrightarrow{0}\)
Ta có: \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}\)
\(\overrightarrow{MN}=\overrightarrow{MD}+\overrightarrow{DC}+\overrightarrow{CN}\)
Do đó: \(2\cdot\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{MD}+\overrightarrow{DC}+\overrightarrow{CN}\)
=>\(2\cdot\overrightarrow{MN}=\overrightarrow{AB}+\overrightarrow{DC}+\left(\overrightarrow{MA}+\overrightarrow{MD}\right)+\left(\overrightarrow{BN}+\overrightarrow{CN}\right)=\overrightarrow{AB}+\overrightarrow{DC}\)
=>\(\overrightarrow{MN}=\frac12\left(\overrightarrow{AB}+\overrightarrow{DC}\right)\)
b:
I là trung điểm của MN
=>\(\overrightarrow{IM}=\overrightarrow{NI}\)
=>\(\overrightarrow{IM}=-\overrightarrow{IN}\)
=>\(\overrightarrow{IM}+\overrightarrow{IN}=\overrightarrow{0}\)
\(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}\)
\(=\overrightarrow{IM}+\overrightarrow{MA}+\overrightarrow{IM}+\overrightarrow{MD}+\overrightarrow{IN}+\overrightarrow{NB}+\overrightarrow{IN}+\overrightarrow{NC}\)
\(=2\cdot\overrightarrow{IM}+\left(\overrightarrow{MA}+\overrightarrow{MD}\right)+2\cdot\overrightarrow{IN}+\left(\overrightarrow{NB}+\overrightarrow{NC}\right)\)
\(=2\cdot\overrightarrow{IM}+2\cdot\overrightarrow{IN}=2\left(\overrightarrow{IM}+\overrightarrow{IN}\right)=\overrightarrow{0}\)
