Đáp án:

Giải thích các bước giải:

Bài 1:

1. a) CM:$\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$

$\overrightarrow{IA}+\overrightarrow{IB}=2\overrightarrow{IM}$ ( Vì $M$ là trung điểm $AB$ )

$\overrightarrow{IC}+\overrightarrow{ID}=2\overrightarrow{IN}$ ( Vì $N$ là trung điểm $CD$ )

$\to \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2\overrightarrow{IM}+2\overrightarrow{IN}$

$\to \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2\left( \overrightarrow{IM}+\overrightarrow{IN} \right)$

$\to \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2.\overrightarrow{0}$ ( Vì $I$ là trung điểm $MN$ )

$\to \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}$

1. b) $\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED}=4\overrightarrow{EI}$

$\overrightarrow{EA}=\overrightarrow{EI}+\overrightarrow{IA}$

$\overrightarrow{EB}=\overrightarrow{EI}+\overrightarrow{IB}$

$\overrightarrow{EC}=\overrightarrow{EI}+\overrightarrow{IC}$

$\overrightarrow{ED}=\overrightarrow{EI}+\overrightarrow{ID}$

$\to \overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED}=4\overrightarrow{EI}+\left( \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID} \right)$

$\to \overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED}=4\overrightarrow{EI}$

Bài 2:

$\bullet \,\,\,\,\,\overrightarrow{PA}=2\overrightarrow{PD}$

$\to \overrightarrow{MA}-\overrightarrow{MP}=2\overrightarrow{MD}-2\overrightarrow{MP}$

$\to \overrightarrow{MP}=2\overrightarrow{MD}-\overrightarrow{MA}\,\,\,\left( 1 \right)$

$\bullet \,\,\,\,\,\overrightarrow{QB}=2\overrightarrow{QC}$

$\to \overrightarrow{MB}-\overrightarrow{MQ}=2\overrightarrow{MC}-2\overrightarrow{MQ}$

$\to \overrightarrow{MQ}=2\overrightarrow{MC}-\overrightarrow{MB}\,\,\,\left( 2 \right)$

Lấy $\left( 1 \right)+\left( 2 \right)$, ta được:

$\overrightarrow{MP}+\overrightarrow{MQ}=2\overrightarrow{MD}-\overrightarrow{MA}+2\overrightarrow{MC}-\overrightarrow{MB}$

$\overrightarrow{MP}+\overrightarrow{MQ}=2\left( \overrightarrow{MD}+\overrightarrow{MC} \right)-\left( \overrightarrow{MA}+\overrightarrow{MB} \right)$

$\overrightarrow{MP}+\overrightarrow{MQ}=2\left( \overrightarrow{MD}+\overrightarrow{MC} \right)\,\,\,\left( 3 \right)$

$\bullet \,\,\,\,\,\overrightarrow{MN}=\frac{1}{2}\left( \overrightarrow{MD}+\overrightarrow{MC} \right)$ ( Vì $N$ là trung điểm $CD$ )

$\to 4\overrightarrow{MN}=2\left( \overrightarrow{MD}+\overrightarrow{MC} \right)\,\,\,\left( 4 \right)$

Từ $\left( 3 \right)$ và $\left( 4 \right)$, ta được:

$4\overrightarrow{MN}=\overrightarrow{MP}+\overrightarrow{MQ}$

$\overrightarrow{MN}=\frac{1}{4}\overrightarrow{MP}+\frac{1}{4}\overrightarrow{MQ}$

Vậy $MNPQ$ đồng phẳng