In triangle $ABC$ we have $M$ the middle of $[BC]$ .How to demonstrate that $vec(AB)+vec(AC)=2vec(AM)$ ?

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Answer 1 · 2017-04-29T14:00:37

See proof below

We apply the sum of 2 vectors ( In France, it is called the "relation of Chasles")

$vec(AB)=vec(AC)+ vec(CB)$

Here

$LHS=vec(AB)+vec(AC)$

$=(vec(AM)+vec(MB))+(vec(AM)+vec(MC))$

As $M$ is the midpoint of $BC$

$vec(BM)=vec(MC)$

$vec(BM)=-vec(MB)=-vec(MC)$

Therefore,

$LHS=vec(AM)-vec(MC)+vec(AM)+vec(MC)$

$=vec(AM)+vec(AM)$

$=2vec(AM)$

$=RHS$

$QED$

In triangle ABCABC we have MM the middle of [BC][BC].How to demonstrate that vec(AB)+vec(AC)=2vec(AM)vec(AB)+vec(AC)=2vec(AM)?