I am not understanding completely the last affermation of the following proof of the converse of Menelaus's Theorem.
In the detail, the author, after having proven in general Menelaus's Theorem for general points $X,Y$ and $Z$ (which lie respectively on sides $BC,CA$ and $AB$, which are extended if necessary), he now proves its converse as follows: (the proof is from AOPS II book)
Moving to our proof, we will show that $YZ,BC$ intersect at the point X which satisfies Menelaus's Theorem. Let this intersection point be $X'$, where $X$ is a point on $BC$ that, along with $Y$ and $Z$ on $AC$ and $AB$ respectively, satisfies $$ \frac {BX \cdot CY \cdot AZ}{ CX \cdot AY \cdot BZ}=1 \tag 1$$
Since $Y$ and $Z$ are on $AC$ and $AB$, and $X'$ is on both $YZ $ and $BC$, we also have: $$\frac{BX' \cdot CY \cdot AZ}{CX' \cdot AY \cdot BZ}=1$$
Combining these we find $$\frac{BX}{CX}=\frac{BX'}{CX'} $$
If these ratios both equal $1$, then we are done, since the above equality will then imply that $X$ and $X'$ are both the midpoint of $BC$. If the costand ratio is not $1$, then we have another problem. (...)
My question is about why the last affermation is true.What are the cases where,given that $\cfrac{BX}{CX}=c$ where $c \ne 1$ and $c \gt 0$, we have multiple points $X'$ on $BC$ such that $\cfrac{BX'}{CX'}=c$ ?
Any help is appreciated.
This can happen if we don't specify whether $X$ is inside or outside interval $\overline{BC}$. If we take two points $X_1,X_2$ such that $X_2$ is between $B,C$ and $B$ is between $X_1,C$ such that $X_1B=BC,X_2B=BC/3$ then we have $X_1C/X_1B=2=X_2C/X_2B$.