Proving the converse of Menelaus’s Theorem

Question

I want to prove the following:

If $\frac{BX}{CX}$$\frac{CY}{AY}$$\frac{AZ}{BZ}=1$, then X, Y and Z are collinear. I have attached what I have got so far, however I am struggling to understand the details for the final part.

My issue comes with the next part, which is how to prove the result when the ratios don’t equal one.

The hint is “Prove that there is only one point $X’$ such that $\frac{BX’}{CX’}=c$ where $c$ is any real number besides $1$”

However, I am not completely sure how to prove this result or how it would lead to the desired outcome. Any guidance would be much appreciated.

Answer 1

$$BX:CX=BX':CX'$$ $$(BX-CX):CX=(BX'-CX'):CX'$$ But $BX-CX=BX'-CX'=BC$ (provided both $X,X'$ are on the right with respect to $C$...), therefore: $$BC:CX=BC:CX' \Rightarrow CX=CX' \Rightarrow X\equiv X'$$

Answer 2

$$\vec{ZY}=-\vec{AZ}+\vec{AY}=-\frac{AZ}{AB}\vec{AB}+\frac{AY}{AC}\vec{AC}=-\frac{AZ}{AZ+BZ}\vec{AB}+\frac{AY}{AY+CY}\vec{AC}.$$ Also, $$\vec{ZX}=\vec{ZB}+\vec{BX}=\frac{BZ}{AZ+BZ}\vec{AB}+\frac{BX}{BX-CX}\vec{BC}=$$ $$=\frac{BZ}{AZ+BZ}\vec{AB}+\frac{BX}{BX-CX}\left(-\vec{AB}+\vec{AC}\right)=$$ $$=\left(\frac{BZ}{AZ+BZ}-\frac{BX}{BX-CX}\right)\vec{AB}+\frac{BX}{BX-CX}\vec{AC}=$$ $$=-\frac{BX\cdot AZ+BZ\cdot CX}{(AZ+BZ)(BX-CX)}\vec{AB}+\frac{BX}{BX-CX}\vec{AC}.$$ Id est, it's enough to prove that $$\vec{ZY}||\vec{ZX}$$ or $$\frac{-\frac{BX\cdot AZ+BZ\cdot CX}{(AZ+BZ)(BX-CX)}}{-\frac{AZ}{AZ+BZ}}=\frac{\frac{BX}{BX-CX}}{\frac{AY}{AY+CY}}$$ or $$\frac{BX\cdot AZ+BZ\cdot CX}{AZ}=\frac{BX(AY+CY)}{AY}$$ or $$AY\cdot BZ\cdot CX=AZ\cdot BX\cdot CY$$ and we are done!