Theo quy tắc ba điểm, ta có:
−−→AD+−−→BE+−−→CF=(−−→AE+−−→ED)+(−−→BF+−−→FE)+(−−→CD+−−→DF)=−−→AE+−−→BF+−−→CD+(−−→FE+−−→ED+−−→DF)=−−→AE+−−→BF+−−→CD+(−−→FD+−−→DF)=−−→AE+−−→BF+−−→CD+→0=−−→AE+−−→BF+−−→CD
Tương tự, ta có:
−−→AD+−−→BE+−−→CF=(−−→AF+−−→FD)+(−−→BD+−−→DE)+(−−→CE+−−→EF)=−−→AF+−−→BD+−−→CE+(−−→FD+−−→DE+−−→EF)=−−→AF+−−→BD+−−→CE+(−−→FE+−−→EF)=−−→AF+−−→BD+−−→CE+→0=−−→AF+−−→BD+−−→CE
Vậy −−→AD+−−→BE+−−→CF
=−−→AE+−−→BF+−−→CD=−−→AF+−−→BD+−−→CE
-- Mod Toán 10 HỌC247