Menelaus' theorem

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Menelaus' theorem, case 1: line DEF passes inside triangle ABC

Menelaus' theorem, attributed to Menelaus of Alexandria, is a theorem about triangles in plane geometry. Given points A, B, C that form triangle ABC, and points D, E, F that lie on lines BC, AC, AB, then the theorem states that D, E, F are collinear if and only if:

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = -1$

In this equation, $A B$ , etc., represent measurements of line segments that are allowed negative values. For example, the fraction $A F / F B$ must be defined as having positive value only when line DEF intersects side AB, and similarly for the other fractions. There is a long-running joke amongst mathematicians that, if this theorem is used to solve a problem, then the 'wrong theorem' was being used (implying that Ceva's theorem should have been used instead).

[edit] Proof

Menelaus' theorem, case 2: line DEF is entirely outside triangle ABC

This is one of many proofs for this theorem. The sign of the left-hand side of the theorem's equation can be checked. Line DEF must intersect the sides of triangle ABC an even number of times—either twice altogether, if it passes into the triangle and out again (upper diagram), or not at all, if it misses the triangle (lower diagram) (Pasch's axiom). Hence there are an odd number of negative terms, and the total product is negative.

Next, the magnitude can be checked. Construct line segments that connect line DEF perpendicularly with vertices A, B, and C. With DEF as the base, let the altitudes of A, B, and C be a, b, and c. By similar triangles, the absolute value of the left-hand side of the theorem simplifies to:

$\left| \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \right| = 1$

Furthermore, we can prove by contradiction that if the theorem's equation holds, then D, E, F must be collinear. Let there be a new point F' on AB different from F, and define the measurements of AF, AF', and AB as n, n', and s. Suppose that F' also satisfies the equation. Then the fractions have equal value:

$\frac{AF}{FB} = \frac{AF'}{F'B}$

$\frac{n}{s - n} = \frac{n'}{s - n'}$

which simplifies to n = n'. This proves that only one point on line AB can satisfy the equation. With D and E fixed, if F satisfies the equation, then it must be the point collinear with D and E. For reasons of symmetry, the same thing can be said about points D and E.

Ptolemy used Menelaus' theorem as the basis for his spherical trigonometry in the Almagest.