The point $X$ divides $AB$ in the ratio $\alpha$, $Y$ divides $BC$ in the ratio $\beta$ and $Z$ divides $CA$ in the ratio $\gamma$. Write $CX,AY,BZ$ in terms of $CA,CB,\alpha, \beta, \gamma$.
I did the following:
I can obtain the points of $AB$ with $CA+x(CB-CA)$ and then this would represent the vector $CX$ for an arbitrary $X\in AB$.
To express the ratio, I took the size of the vector $(CB-CA)$ and wrote it in two parts: One that express the size of the vector $AX$ and another that express the size of the vector $XB$:
- $|AX|=|x(CB-CA)|$;
- $|BX|=|CB-CA|-|x(CB-CA)|$;
Then the ratio is:
$$|AX|=r|XB|$$
$$r=\cfrac{|AX|}{|XB|}=\cfrac{|x(CB-CA)|}{|CB-CA|-|x(CB-CA)|}$$
But I'm don't feel it's over and I don't know how to proceed from here. So, Is my reasoning correct until now? If yes, how can I proceed?
Assume that $X$ is between $A$ and $b$, we have $$\alpha = \frac{AX}{BX}$$, then $$\vec{AX} + \alpha\vec{BX} = \vec{0}.$$
So, $$\vec{CA} + \alpha\vec{CB} = (1+\alpha)\vec{CX}.$$
Then, we get $$(1+\alpha)|CX| = \sqrt{|CA|^2 + \alpha^2|CB|^2 + 2\alpha|CA||CB| \cos(\vec{CA},\vec{CB})}$$
Similarly, we have $$(1+\beta)|AY| = \sqrt{|AB|^2 + \beta^2|AC|^2 + 2\beta|CA||CB| \cos(\vec{AB},\vec{AC})}$$
$$(1+\gamma)|BZ| = \sqrt{|BA|^2 + \gamma^2|BC|^2 + 2\gamma|BA||BC| \cos(\vec{BA},\vec{BC})}$$