We have ABCABC a scalene triangle, and a point MM in plane of this triangle. How to prove that vec(AB)*vec(CM)+vec(AC)*vec(MB)+vec(AM)*vec(BC) = 0ABCM+ACMB+AMBC=0?

2 Answers
Jul 10, 2018

Please see the proof below

Explanation:

Apply Chasles' Relation

vec(AB)*vec(CM)=(vec(AM)+vec(MB))vec(CM)=vec(AM)*vec(CM)+vec(MB)*vec(CM)ABCM=(AM+MB)CM=AMCM+MBCM

vec(AC)*vec(MB)=(vec(AM)+vec(MC))vec(MB)=vec(AM)*vec(MB)+vec(MC)*vec(MB)ACMB=(AM+MC)MB=AMMB+MCMB

vec(BC)*vec(AM)=(vec(BM)+vec(MC))vec(AM)=vec(BM)*vec(AM)+vec(MC)*vec(AM)BCAM=(BM+MC)AM=BMAM+MCAM

But,

vec(CM)=-vec(MC)CM=MC

vec(MB)=-vec(BM)MB=BM

Therefore, Adding the first 33 equations

(vec(AB)*vec(CM)+vec(AC)*vec(MB)+vec(BC)*vec(AM))(ABCM+ACMB+BCAM)

=vec(AM)*vec(CM)+vec(MB)*vec(CM)+vec(AM)*vec(MB)+vec(MC)*vec(MB)+vec(BM)*vec(AM)+vec(MC)*vec(AM)=AMCM+MBCM+AMMB+MCMB+BMAM+MCAM

=vec(AM)*vec(CM)-vec(AM)*vec(CM)+vec(MB)*vec(CM)-vec(MB)*vec(CM)+vec(AM)*vec(MB)-vec(AM)*vec(MB)=AMCMAMCM+MBCMMBCM+AMMBAMMB

=0=0

Jul 10, 2018

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We have ABCABC a scalene triangle, and a point MM in the plane of this triangle. We are to prove that vec(AB)*vec(CM)+vec(AC)*vec(MB)+vec(AM)*vec(BC) = 0ABCM+ACMB+AMBC=0

Now by triangle law we have

For DeltaABC,vec(AB)=vec(AC)+vec(CB)....[1]

For DeltaBMC,vec(CM)=vec(CB)-vec(MB).. .[2]

For DeltaAMC,vec(CM)=vec(AM)-vec(AC).. .[3]

Now using [1] we get

vec(AB)*vec(CM)=vec(AC)*vec(CM)+vec(CB)*vec(CM)

=>vec(AB)*vec(CM)=vec(AC)*vec(CM)-vec(BC)*vec(CM)

=>vec(AB)*vec(CM)=vec(AC)*(vec(CB)-vec(MB))-vec(BC)*(vec(AM)-vec(AC))

=>vec(AB)*vec(CM)=-vec(AC)*vec(BC)-vec(AC)*vec(MB)-vec(BC)*vec(AM)+vec(BC)vec(AC)

=>vec(AB)*vec(CM)+vec(AC)*vec(MB)+vec(AM)*vec(BC) = 0