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Question-94191




Question Number 94191 by oustmuchiya@gmail.com last updated on 17/May/20
Answered by mr W last updated on 17/May/20
AD^(→) =c+(a/2)  BE^(→) =a+(b/2)  CF^(→) =b+(c/2)  AD^(→) +BE^(→) +CF^(→) =  (c+(a/2))+(a+(b/2))+(b+(c/2))=(3/2)(a+b+c)=0
$$\overset{\rightarrow} {{AD}}=\boldsymbol{{c}}+\frac{\boldsymbol{{a}}}{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{BE}}=\boldsymbol{{a}}+\frac{\boldsymbol{{b}}}{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{CF}}=\boldsymbol{{b}}+\frac{\boldsymbol{{c}}}{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{AD}}+\overset{\rightarrow} {{BE}}+\overset{\rightarrow} {{CF}}= \\ $$$$\left(\boldsymbol{{c}}+\frac{\boldsymbol{{a}}}{\mathrm{2}}\right)+\left(\boldsymbol{{a}}+\frac{\boldsymbol{{b}}}{\mathrm{2}}\right)+\left(\boldsymbol{{b}}+\frac{\boldsymbol{{c}}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)=\mathrm{0} \\ $$
Commented by mr W last updated on 17/May/20

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