Ta có: $\overrightarrow{MB}+\overrightarrow{BN}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC} (1)$Tương tự: $\overrightarrow{PQ}=\overrightarrow{PD}+\overrightarrow{DQ}=\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{DA} (2)$Cộng $(1),(2)$ vế với vế ta có:$\overrightarrow{MN}+\overrightarrow{PQ}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA})=\overrightarrow{0} $Suy ra: $\overrightarrow{MN}=-\overrightarrow{PQ}=\overrightarrow{QP}$. Vậy $MNPQ$ là hình bình hành.
Ta có: $\overrightarrow{M
N}=\overrightarrow{MB}+\overrightarrow{BN}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC} (1)$Tương tự: $\overrightarrow{PQ}=\overrightarrow{PD}+\overrightarrow{DQ}=\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{DA} (2)$Cộng $(1),(2)$ vế với vế ta có:$\overrightarrow{MN}+\overrightarrow{PQ}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA})=\overrightarrow{0} $Suy ra: $\overrightarrow{MN}=-\overrightarrow{PQ}=\overrightarrow{QP}$. Vậy $MNPQ$ là hình bình hành.