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

1 Answer
Apr 29, 2017

See proof below

Explanation:

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