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Question Number 140941 by mr W last updated on 14/May/21

Commented by mr W last updated on 14/May/21

$$seeQ140735$$

Commented by mr W last updated on 14/May/21

$$\overrightarrow{OA}=\mathit{a}$$$$\overrightarrow{OB}=\mathit{b}$$$$\overrightarrow{OC}=\mathit{c}$$$$\overrightarrow{OD}=\frac{\mathit{b}+\mathit{c}}{2}$$$$\overrightarrow{OE}=\frac{\mathit{a}+\mathit{c}}{2}$$$$\overrightarrow{OF}=\frac{\mathit{a}+\mathit{b}}{2}$$$$\overrightarrow{AD}=\frac{\mathit{b}+\mathit{c}}{2}-\mathit{a}$$$$\overrightarrow{BE}=\frac{\mathit{a}+\mathit{c}}{2}-\mathit{b}$$$$sayADandBEintersectatG$$$$with\frac{AG}{AD}=\lambda ,\frac{BG}{BE}=\mu$$$$\overrightarrow{OG}=\mathit{a}+\lambda (\frac{\mathit{b}+\mathit{c}}{2}-\mathit{a})=\mathit{b}+\mu (\frac{\mathit{a}+\mathit{c}}{2}-\mathit{b})$$$$(1-\lambda -\frac{\mu }{2})\mathit{a}+(\frac{\lambda }{2}-1+\mu )\mathit{b}+(\frac{\lambda }{2}-\frac{\mu }{2})\mathit{c}=0$$$$\frac{\lambda }{2}-\frac{\mu }{2}=0\Rightarrow \mu =\lambda$$$$1-\lambda -\frac{\mu }{2}=0\Rightarrow 1-\frac{3\lambda }{2}=0\Rightarrow \lambda =\frac{2}{3}$$$$\frac{\lambda }{2}-1+\mu =0\Rightarrow \frac{3\mu }{2}-1=0\Rightarrow \mu =\frac{2}{3}$$$$similarlywegetthatCFandAD$$$$intersectat{G}^{\prime }with\frac{{AG}^{\prime }}{AD}=\frac{2}{3},that$$$$means{G}^{\prime }isthesamepointasG,$$$$i.e.allthreemediansintersectat$$$$thesamepointGwith$$$$\frac{AG}{AD}=\frac{BG}{BE}=\frac{CG}{CF}=\frac{2}{3}$$

Answered by ajfour last updated on 14/May/21

$$AB=\overline{c}\left(letAorigin\right)$$$$AC=\overline{b}$$$$lineCFr=b+\lambda (c-2b)$$$$lineADr=\mu (b+c)$$$$intersection$$$$\lambda =\mu and1-2\lambda =\mu$$$$\Rightarrow \lambda =\mu =\frac{1}{3}$$$$\Rightarrow r_G=\frac{b+c}{3}$$$$lineBEr=c+\rho (b-2c)$$$$letscheckif{r}_{G}liesonit$$$$\frac{b+c}{3}=c+\rho (b-2c)$$$$\Rightarrow {\rho }_1=\frac{1}{3}and{\rho }_2=\frac{1}{3}$$$$henceaffirmed.$$

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