# Menelaus's Theorem

## Theorem

Let $ABC$ be a triangle.

Let points $D, E, F$ lie on lines $BC, AC, AB$ respectively (produced if necessary).

Then $D, E$ and $F$ are collinear if and only if:

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

In the above, the line segments $AF, BD, EA$ are determined to have negative length if they lie outside the line segments $AB, BC, CA$.

## Proof

### Necessary Condition

First we check that the equation works out negative.

From Pasch's Axiom, the line $DEF$ must intersect either two sides of $\triangle ABC$:

or none of them:

That means there is an odd number of negative contributions to the product.

Hence the equation works out to be negative.

Next, the magnitude can be checked.

The method is illustrated using the second of the above options; the first works the same.

Construct perpendiculars $AD$, $CH$ and $BI$ from $A$, $B$ and $C$ to $DEF$.

The triangles $\triangle AEG$, $\triangle CEH$ are similar.

The triangles $\triangle BDI$, $\triangle CDH$ are also similar.

The triangles $\triangle BFI$, $\triangle AFG$ are also similar.

Hence:

- $\dfrac {CE} {EA} = \dfrac {CH} {AG}, \dfrac {BD} {DC} = \dfrac {BI} {CH}, \dfrac {AF} {FB} = \dfrac {AG} {BI}$

Thus:

- $\size {\dfrac {AF} {FB} \cdot \dfrac{BD} {DC} \cdot \dfrac{CE} {EA} } = \size {\dfrac {AG} {BI} \cdot \dfrac {BI} {CH} \cdot \dfrac {CH} {AG} } = 1$

Hence the result.

$\Box$

### Sufficient Condition

Suppose $\dfrac {AF} {FB} \cdot \dfrac {BD} {DC} \cdot \dfrac {CE} {EA} = -1$.

Let $F'$ be a point on $AB$ distinct from $F$.

Let us define the measurements of $AF$, $AF'$, $AB$ as $n, n', s$ respectively.

Without loss of generality, suppose that $F'$ also satisfies the equation.

Then:

- $\dfrac {AF} {FB} = \dfrac {AF'} {F'B}$

which means:

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

and so $n = n'$ and so $F = F'$.

So only one point on line $AB$ can satisfy the equation.

Let us fix $D$ and $E$.

Then if $F$ satisfies the equation, then it must be the point collinear with $D$ and $E$.

By a similar argument, the same applies to $D$ and $E$.

Hence the result.

$\blacksquare$

## Also see

- Ceva's Theorem, to which this one is related.

## Source of Name

This entry was named for Menelaus of Alexandria.

## Sources

- 1989: Ephraim J. Borowski and Jonathan M. Borwein:
*Dictionary of Mathematics*... (previous) ... (next): Entry:**Menelaus' theorem** - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**Menelaus' theorem** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Menelaus' theorem** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**Menelaus' Theorem**