Ceva and Menelaus

Here is a result I worked out using the  theorems of Ceva and Menelaus.
The points P, Q and R lie on the sides BC, CA and AB of triangle ABC respectively (not necessarily in their interiors), such that they are not collinear. RQ and BC meet at X; RP and AC meet at Y; and QP and AB meet at Z. Then, AP, BQ and CR are concurrent if and only if X, Y and Z are collinear.

1. If AP, BQ and CR are concurrent, then X, Y and Z are collinear.

Proof: The fact that P, Q and R are not collinear prevents degenerate diagrams (where X, Y and Z coincide with P, Q and R). So, applying Ceva to triangle ABC, we obtain:
$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=1$
Treating RY as a transversal, we obtain, by Menelaus,
$\frac{AR}{RB}\frac{BP}{PC}\frac{CY}{YA}=-1$
Treating XQ as a transversal, we obtain, by Menelaus,
$\frac{AR}{RB}\frac{BX}{XC}\frac{CQ}{QA}=-1$
Treating ZQ as a transversal, we obtain, by Menelaus,
$\frac{AZ}{ZB}\frac{BP}{PC}\frac{CQ}{QA}=-1$
Dividing the first equation by the second yields:
$\frac{CQ}{QA}=-\frac{CY}{YA}$
Dividing the first equation by the third yields:
$\frac{BP}{PC}=-\frac{BX}{XC}$
Dividing the first equation by the fourth yields:
$\frac{AR}{RB}=-\frac{AZ}{ZB}$
Substituting back into the first equation,
$(-\frac{CY}{YA})(-\frac{AZ}{ZB})(-\frac{BX}{XC})=1$
$\frac{CY}{YA}\frac{AZ}{ZB}\frac{BX}{XC}=-1$
So by the converse of Menelaus, XZY must be a transversal and hence a straight line.

2. If X, Y and Z are collinear, then AP, BQ and CR are concurrent.

Proof: By Menelaus,
$\frac{CY}{YA}\frac{AZ}{ZB}\frac{BX}{XC}=-1$
As before, treating RY as a transversal, we obtain, by Menelaus,
$\frac{AR}{RB}\frac{BP}{PC}\frac{CY}{YA}=-1$
Treating XQ as a transversal, we obtain, by Menelaus,
$\frac{AR}{RB}\frac{BX}{XC}\frac{CQ}{QA}=-1$
Treating ZQ as a transversal, we obtain, by Menelaus,
$\frac{AZ}{ZB}\frac{BP}{PC}\frac{CQ}{QA}=-1$
Multiplying these last three equations,
$\frac{AR}{RB}\frac{BP}{PC}\frac{CY}{YA}\frac{AR}{RB}\frac{BX}{XC}\frac{CQ}{QA}\frac{AZ}{ZB}\frac{BP}{PC}\frac{CQ}{QA}=-1$
Or,
$\frac{CY}{YA}\frac{AZ}{ZB}\frac{BX}{XC}(\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA})^{2}=-1$
Dividing by the first equation,
$(\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA})^{2}=1$
So
$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=\pm 1$
If
$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=-1$
then by the converse of Menelaus, P, Q and R are collinear. This is not the case, so we reject -1 as a solution.
So
$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=1$
So by the converse of Ceva, AP, BQ and CR are concurrent.