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