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Question Number 16055 by Tinkutara last updated on 17/Jun/17

$$\mathrm{Let}{A}^{\prime },B^{\prime }\mathrm{and}{C}^{\prime }\mathrm{be}\mathrm{the}\mathrm{midpoints}\mathrm{of}$$$$\mathrm{the}\mathrm{sides}BC,CA\mathrm{and}AB\mathrm{of}\mathrm{the}$$$$\mathrm{triangle}ABC.\mathrm{Prove}\mathrm{that}$$$$\overrightarrow{{AA}^{\prime }}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$$

Answered by mrW1 last updated on 17/Jun/17

$$\overrightarrow{{AA}^{\prime }}=\overrightarrow{\mathrm{AB}}+\frac{1}{2}\overrightarrow{\mathrm{BC}}...\left(\mathrm{i}\right)$$$$\overrightarrow{{AA}^{\prime }}=\overrightarrow{\mathrm{AC}}+\frac{1}{2}\overrightarrow{\mathrm{CB}}=\overrightarrow{\mathrm{AC}}-\frac{1}{2}\overrightarrow{\mathrm{BC}}...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right):$$$$2\overrightarrow{{AA}^{\prime }}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}$$$$\overrightarrow{{AA}^{\prime }}=\frac{1}{2}(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}})$$

Commented by Tinkutara last updated on 17/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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