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 iff:

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

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.

We illustrate the method using the second of the above options; the first works the same.

We 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 $\displaystyle \frac {CE} {EA} = \frac {CH} {AG}, \frac {BD} {DC} = \frac {BI} {CH}, \frac {AF} {FB} = \frac {AG} {BI}$.

Thus:

$\displaystyle \left|{\frac{AF} {FB} \cdot \frac{BD} {DC} \cdot \frac{CE} {EA}}\right| = \left|{\frac {AG} {BI} \cdot \frac {BI} {CH} \cdot \frac {CH} {AG}}\right| = 1$.

Hence the result.

Sufficient Condition

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

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

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

Suppose WLOG that $F'$ also satisfies the equation. Then we have:

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

which means

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

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.

$\blacksquare$

See also

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

Source of Name

This entry was named for Menelaus of Alexandria.