Prove that the points A,B,C are collinear with the point $A$ lying between the points $B$ and $C$

Question

Let $A(z_{1}),B(z_{2}),C(z_{3})$ be three points in a plane such that $$z_{1}|z_{2}-z_{3}|-z_{2}|z_{3}-z_{1}|-z_{3}|z_{1}-z_{2}|=0$$ Then prove that if the points A,B,C are collinear then the point $A$ lies between the points $B$ and $C$.

My Attempt

$$z_{1}=\frac{z_{2}|z_{3}-z_{1}|+z_{3}|z_{1}-z_{2}|}{|z_{2}-z_{3}|}$$ Now $$z_{4}=\frac{z_{2}|z_{3}-z_{1}|+z_{3}|z_{1}-z_{2}|}{|z_{3}-z_{1}|+|z_{1}-z_{2}|}$$ is a point on $BC$ and also on angle bisector of the angle $BAC$ .What should be the condition so that $z_{4}$ gets expressed as $z_{1}$

I mean by what reason should $$|z_{2}-z_{3}|=|(z_{2}-z_{1})-(z_{3}-z_{1})|=|z_{3}-z_{1}|+|z_{1}-z_{2}|$$

Answer 1

We acn assume that $A$ at $0$ and $B$ and $C$ are at $w$ and $u$. So we have $u|w|+w|u|=0$ i.e. $${u\over |u|}+{w\over |w|}=0$$ So if $a={u\over |u|}$ then ${w\over |w|}=-a$. So $u=a|u|$ and $w =-a|w|$ which means that $u$ and $w$ are number $a$ multiplied with some real number and thus $u,w$ and $0$ are collinear with $0$ between them.

Answer 2

Let $w_2= z_2-z_1$ and $w_3= z_3-z_1$. Then, the equation

$$z_{1}|z_{2}-z_{3}|-z_{2}|z_{3}-z_{1}|-z_{3}|z_{1}-z_{2}|=0$$ becomes

$$z_{1}|w_{2}-w_{3}|-(w_{2}+z_1)|w_{3}|-(w_{3}+z_1)|w_{2}|=0$$

Rearrange to get

$$z_{1}(|w_{2}-w_{3}|-|w_{2}|-|w_{3}|)=w_{3}|w_{2}| +w_{2}|w_{3}|$$

which shall hold regardless the actual coordinate of $z_1$, implying

$$|w_{2}-w_{3}|-|w_{2}|-|w_{3}|=w_{3}|w_{2}| +w_{2}|w_{3}|=0$$

Thus,$|w_{2}-w_{3}|=|w_{2}|+|w_{3}|$, or

$$|z_{2}-z_{3}|=|z_{3}-z_{1}|+|z_{1}-z_{2}|$$