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:

Treating RY as a transversal, we obtain, by Menelaus,

Treating XQ as a transversal, we obtain, by Menelaus,

Treating ZQ as a transversal, we obtain, by Menelaus,

Dividing the first equation by the second yields:

Dividing the first equation by the third yields:

Dividing the first equation by the fourth yields:

Substituting back into the first equation,

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,

As before, treating RY as a transversal, we obtain, by Menelaus,

Treating XQ as a transversal, we obtain, by Menelaus,

Treating ZQ as a transversal, we obtain, by Menelaus,

Multiplying these last three equations,

Or,

Dividing by the first equation,

So

If

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

So by the converse of Ceva, AP, BQ and CR are concurrent.

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