Consider a plane containing a triangle ABC and a line that intersects each of the extended (to infinity) sides of the triangle. Obviously no line can intersect all three sides of a triangle, so at least one intersection has to be on the extension of a triangle side. Let D be the intersection with line BC, E the intersection with CA, and F the intersection with AB.

The intersection points will divide each side into segments. For instance, the line BC is separated into BD and DC. Treat the segments as signed segments, so that if D lies on the exterior of the triangle, then either BD or DC is negative.

**The Theorem of Menelaus states that (BD/DC)*(CE/EA)*(AF/FB) = -1.** The proof is boring.