How was Menelaus' theorem used here? I have a few questions about Menelaus' theorem. Firstly, I have seen some versions of Menelaus' theorem have a negative sign, i.e. here. My question is is it really necessary to have a negative sign? Because I have seen other forms of Menelaus' theorem without a negative sign, specifically here.
My second question is related to a problem involving Menelaus' theorem shown below with the solution.

How did they conclude that $F,X,C$ are collinear using it? I drew it (as shown below) and they can't possibly be collinear unless $X$ lies on $FC$.

 A: As for the first question, the answer is, in my opinion, yes and no.
For when you apply the Menelaus' theorem itself, you don't need to consider signed lengths. However, when you want to use the converse of Menelaus, it is useful - though not necessary, look below - to consider signed lengths.
What I mean by that - if you consider unsigned lengths, then both Menelaus' theorem and Ceva's theorem have the same conclusion. To differentiate between them, you need to make additional assumption. Namely, for converse of Menelaus', you need to have exactly two or none of the points $D,E,F$ to lie inside line segments determining sides, while for Ceva's exactly one or all of them line inside these line segments.
However, if you consider signed lengths, all trouble instantly goes away - Ceva gives $+1$, Menelaus gives $-1$ and inverse theorems need only to assume the signed product to be $+1$ or $-1$, respectively.
As for second question, you are right - there is a mistake either in the statement of the problem or in the solution. To be more precise, the statement and solution disagree about which of the problems $X,Y,Z$ lie where. To make the solution correct, one needs $X$ to be the intersection of $AD,CF$, $Y$ to be the intersection of $AD,BE$, and $Z$ to be the intersection of $BE,CF$.
A: It seems, there is a typo in the formulation of the problem. If AD and BE  intersect in Y, BE and CF - in Z and CF and AD in X, the proof will fit correctly. (You just need to change your notation in the following way $X \to \hat{Y}, Y \to \hat{Z}, Z \to \hat{X}$, where nee points are with a hat. This new points will fit into the proof of the theorem.
