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Question Number 94191 by oustmuchiya@gmail.com last updated on 17/May/20

Answered by mr W last updated on 17/May/20

$$\overrightarrow{AD}=\mathit{c}+\frac{\mathit{a}}{2}$$$$\overrightarrow{BE}=\mathit{a}+\frac{\mathit{b}}{2}$$$$\overrightarrow{CF}=\mathit{b}+\frac{\mathit{c}}{2}$$$$\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=$$$$(\mathit{c}+\frac{\mathit{a}}{2})+(\mathit{a}+\frac{\mathit{b}}{2})+(\mathit{b}+\frac{\mathit{c}}{2})=\frac{3}{2}(\mathit{a}+\mathit{b}+\mathit{c})=0$$

Commented by mr W last updated on 17/May/20

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