# We have ABC a scalene triangle, and a point M in plane of this triangle. How to prove that vec(AB)*vec(CM)+vec(AC)*vec(MB)+vec(AM)*vec(BC) = 0?

Jul 10, 2018

#### Explanation:

Apply Chasles' Relation

$\vec{A B} \cdot \vec{C M} = \left(\vec{A M} + \vec{M B}\right) \vec{C M} = \vec{A M} \cdot \vec{C M} + \vec{M B} \cdot \vec{C M}$

$\vec{A C} \cdot \vec{M B} = \left(\vec{A M} + \vec{M C}\right) \vec{M B} = \vec{A M} \cdot \vec{M B} + \vec{M C} \cdot \vec{M B}$

$\vec{B C} \cdot \vec{A M} = \left(\vec{B M} + \vec{M C}\right) \vec{A M} = \vec{B M} \cdot \vec{A M} + \vec{M C} \cdot \vec{A M}$

But,

$\vec{C M} = - \vec{M C}$

$\vec{M B} = - \vec{B M}$

Therefore, Adding the first $3$ equations

$\left(\vec{A B} \cdot \vec{C M} + \vec{A C} \cdot \vec{M B} + \vec{B C} \cdot \vec{A M}\right)$

$= \vec{A M} \cdot \vec{C M} + \vec{M B} \cdot \vec{C M} + \vec{A M} \cdot \vec{M B} + \vec{M C} \cdot \vec{M B} + \vec{B M} \cdot \vec{A M} + \vec{M C} \cdot \vec{A M}$

$= \vec{A M} \cdot \vec{C M} - \vec{A M} \cdot \vec{C M} + \vec{M B} \cdot \vec{C M} - \vec{M B} \cdot \vec{C M} + \vec{A M} \cdot \vec{M B} - \vec{A M} \cdot \vec{M B}$

$= 0$

Jul 10, 2018 We have $A B C$ a scalene triangle, and a point $M$ in the plane of this triangle. We are to prove that $\vec{A B} \cdot \vec{C M} + \vec{A C} \cdot \vec{M B} + \vec{A M} \cdot \vec{B C} = 0$

Now by triangle law we have

For $\Delta A B C , \vec{A B} = \vec{A C} + \vec{C B} \ldots . \left[1\right]$

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

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

Now using  we get

$\vec{A B} \cdot \vec{C M} = \vec{A C} \cdot \vec{C M} + \vec{C B} \cdot \vec{C M}$

$\implies \vec{A B} \cdot \vec{C M} = \vec{A C} \cdot \vec{C M} - \vec{B C} \cdot \vec{C M}$

$\implies \vec{A B} \cdot \vec{C M} = \vec{A C} \cdot \left(\vec{C B} - \vec{M B}\right) - \vec{B C} \cdot \left(\vec{A M} - \vec{A C}\right)$

$\implies \vec{A B} \cdot \vec{C M} = - \vec{A C} \cdot \vec{B C} - \vec{A C} \cdot \vec{M B} - \vec{B C} \cdot \vec{A M} + \vec{B C} \vec{A C}$

$\implies \vec{A B} \cdot \vec{C M} + \vec{A C} \cdot \vec{M B} + \vec{A M} \cdot \vec{B C} = 0$