Prove that $X,Y,Z$ are collinear. 
$AD,BE$ and $CF$ are three concurrent lines meeting the sides $BC,CA,AB$ in $D,E,F$. Suppose $EF,FD$ and $DE$ meet $BC,CA,AB$ at $X,Y,Z$. Prove that $B,C$ divide $DX$ harmonically and $X,Y,Z$ are collinear.

What I Tried:- Here is a Figure :-

I have the answer of the first part here, which is understandable :- AD, BE CF are concurrent lines in triangle ABC. EF, FD ,DE meet BC,CA,AB at X,Y,Z
Now the second part I think, is also easy, which I need to show by the Converse of Menelaus Theorem, but in these type of problems I usually get confused with the side-lengths. As far as I could decide, to show $X,Y,Z$ are collinear I need to show :=
$$\frac{AY}{CY} * \frac{BZ}{AZ} * \frac{CX}{AX} = -1.$$
Is this correct? Someone Please Confirm.
I also have no idea how to show this claim. I tried in some particular ways, but it didn't work.
Can someone guide me through this problem?
 A: Points $X$, $Y$ and $Z$ are on side $BC$, $CA$ and $AB$ respectively. By Menelaus' Theorem, they are collinear iff $\frac {BX}{XC}\cdot \frac {CY}{YA} \cdot \frac {AZ}{BZ}=1$
Applying Ceva's Theorem on $\triangle ABC$ gives,
$\frac {BD}{DC}\cdot \frac {CE}{EA} \cdot \frac {AF}{FB}=1$
Applying Menelaus'  theorem on $\triangle ABC$ considering $FEX$ as the transversal gives,
$\frac {BD}{DC}\cdot \frac {CE}{EA} \cdot \frac {AZ}{BZ}=1$
Similarly,  applying Menelaus'  theorem on $\triangle ABC$ considering $FDY$ and $DEZ$ as transversal respectively gives,
$\frac {CE}{EA}\cdot \frac {AF}{FB} \cdot \frac {BX}{XC}=1$
$\frac {AF}{FB}\cdot \frac {BD}{DC} \cdot \frac {CY}{YA}=1$
Multiplying these three equations and then following Ceva's  Theorem gives the desired result.
$\frac {CE}{EA}\cdot \frac {AF}{FB} \cdot \frac {BX}{XC}=1$
$\Rightarrow \frac {BX}{XC}=\frac {BD}{DC}$ by Ceva's Theorem and thereafter points $D$,  $X$ divide $BC$ harmonically. So do points $F$, $Z$ and $E$, $Y$ to sides $AB$ and $CA$ respectively.
