cho tam giác ABC và K bất kì cmr
[tex]S_{KBC}*\overrightarrow{KA}+ S_{KCA}*\overrightarrow{KB}+ S_{KAB}*\overrightarrow{KC}=\overrightarrow{0}[/tex]
Kẻ AH_|_BM, BJ_|_AM, CI_|_BM
Dựng hbh $KECF$ , [tex]E\in AM, \ F\in BM[/tex]
Theo quy tắc hbh có:
[tex]\overrightarrow{KC}=\overrightarrow{KE}+\overrightarrow{KF}=\frac{-KE}{KA}\overrightarrow{KA}+\frac{-KF}{KB}\overrightarrow{KB}[/tex]
Ta có:
[tex]\frac{KE}{KA}=\frac{CF}{KA}=\frac{CI}{AH}=\frac{S_{KBC}}{S_{KAB}} \ (1) \\ \frac{KF}{KB}=\frac{EC}{KB}=\frac{S_{KAC}}{S_{KAB}} \ (2)[/tex]
Từ (1) và (2) [tex]\Rightarrow \overrightarrow{KC}=\frac{-S_{KBC}}{S_{KAB}}\overrightarrow{KA}+\frac{-S_{KAC}}{S_{KAB}}\overrightarrow{KB} \\ \Rightarrow S_{KAB}\overrightarrow{MC}+S_{KAC}\overrightarrow{KA}+S_{KAC}\overrightarrow{KB}=\overrightarrow{0}[/tex]