Let-A-B-and-C-be-the-midpoints-of-the-sides-BC-CA-and-AB-of-the-triangle-ABC-Prove-that-AA-1-2-AB-AC-

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

$$\mathrm{Let}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoints}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:{ABC}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\overset{\rightarrow} {{AA}'}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\overset{\rightarrow} {{AB}}\:+\:\overset{\rightarrow} {{AC}}\right) \\ $$

Answered by mrW1 last updated on 17/Jun/17

$$\overset{\rightarrow} {{AA}'}\:=\:\overset{\rightarrow} {\mathrm{AB}}+\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {\mathrm{BC}}\:…\left(\mathrm{i}\right) \\ $$$$\overset{\rightarrow} {{AA}'}\:=\:\overset{\rightarrow} {\mathrm{AC}}+\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {\mathrm{CB}}=\overset{\rightarrow} {\mathrm{AC}}−\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {\mathrm{BC}}\:…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right): \\ $$$$\mathrm{2}\overset{\rightarrow} {{AA}'}\:=\:\overset{\rightarrow} {\mathrm{AB}}+\overset{\rightarrow} {\mathrm{AC}} \\ $$$$\overset{\rightarrow} {{AA}'}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\overset{\rightarrow} {\mathrm{AB}}+\overset{\rightarrow} {\mathrm{AC}}\right) \\ $$

Commented by Tinkutara last updated on 17/Jun/17

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

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