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Question

Cho tu giac $ABCD. G$ la trong tam cua tam giac $ABD , I\in GC $sao cho $IC=3IG$ . Chung minh rang voi moi diem M ta luon co:$\overrightarrow{MA}$+$\overrightarrow{MB}$+$\overrightarrow{MC}$+$\overrightarrow{MD}$=$4\overrightarrow{MI}$

Dép Lê Con Nhà Quê · Answer 1 · 10/31/2013 9:18:10 AM

$\overrightarrow{MA} +\overrightarrow{MB} +\overrightarrow{MC} +\overrightarrow{MD}= 4\overrightarrow{MI} +\overrightarrow{IA} +\overrightarrow{IB} +\overrightarrow{IC}+ \overrightarrow{ID}$

$=4\overrightarrow{MI} +4\overrightarrow{IG} +\overrightarrow{GA} +\overrightarrow{GB}+ \overrightarrow{GC} +\overrightarrow{GD} =4\overrightarrow{MI} +4\overrightarrow{IG} +\overrightarrow{GC} \ (1)$

Theo bài ra $IC = 3IG \Rightarrow 3\overrightarrow{IG}=- \overrightarrow{IC}$ thế lên $(1)$ ta có

$(1) = \overrightarrow{MA} +\overrightarrow{MB} +\overrightarrow{MC} +\overrightarrow{MD}= 4\overrightarrow{MI} +\overrightarrow{IA} +\overrightarrow{IB} +\overrightarrow{IC}+ \overrightarrow{ID}$

$=4\overrightarrow{MI} +4\overrightarrow{IG} +\overrightarrow{GA} +\overrightarrow{GB}+ \overrightarrow{GC} +\overrightarrow{GD} =4\overrightarrow{MI} +\overrightarrow{IG} -\overrightarrow{IC}+\overrightarrow{GC} $

$=4\overrightarrow{MI} +\overrightarrow{IC} -\overrightarrow{IC}=4\overrightarrow{MI}$