# Menelaus' Theorem

**Menelaus' theorem** relates ratios obtained by a line cutting the sides of a triangle. The converse of the theorem (i.e. three points on a triangle are collinear if and only if they satisfy certain criteria) is also true and is extremely powerful in proving that three points are collinear.

**Ceva's theorem** is essentially the counterpart of this theorem and can be used to prove three lines are concurrent at a single point. Both theorems possess similar structures and are widely applicable in various geometry problem types.

## Theorem

Menelaus' theoremstates that if a line intersects $\triangle ABC$ or extended sides at points $D$, $E$, and $F$, the following statement holds:$\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=1.$

Converse of Menelaus' Theorem: Suppose three points $D,E,F$ are on sides (or extension) $AB,BC,AC$ respectively, such that $1$ or $3$ of them are in the extensions of the sides. Then points $D,E,F$ are collinear if and only if$\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=1.$

The **Converse of Menelaus' Theorem** is very powerful in proving that three points are collinear, especially in Olympiad problems.

Construct lines $AA'$, $BB'$, and $CC'$ that are perpendicular to the yellow line.

Now, since $\triangle AA'D\sim \triangle BB'D$, $\frac{AD}{DB}=\frac{AA'}{BB'}.$ Since $\triangle AA'F\sim \triangle CC'F$, $\frac{CF}{FA}=\frac{CC'}{AA'}.$ Since $\triangle BB'E\sim \triangle CC'E$, $\frac{BE}{EC}=\frac{BB'}{CC'}.$

Hence, $\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=\frac{AA'}{BB'}\times\frac{BB'}{CC'}\times\frac{CC'}{AA'}=1.\ _\square$

**Note:**

- The equality still holds even when the yellow line does not intersect the triangle at all (that means the yellow line intersects at the extended parts of all sides of the triangle).
- If the yellow line intersects one of the vertices of the triangle, then a 0 will appear in the denominator of the equation, which is undefined; to solve this problem, the Menelaus' theorem could also be rewritten as $AD\times BE\times CF=DB\times EC\times FA$.

Menelaus drew triangle $ABC$ with $BC = 13$ before crossing two red lines $BD = 10$ and $CE = 15,$ both intersecting at point $P$ and reaching the triangle's sides at points $D$ and $E,$ respectively, as shown above.

**Menelaus**: Mark this, lad. Point $P$ does not only divide all the red segments into integer lengths, but points $D$ and $E$ also divide the triangle's sides into integer lengths.

**Pupil**: O, so true, master! Any length between any two of those points is always a whole integer!

**Menelaus**: Then thou shalt tell me. What is the perimeter of triangle $ABC?$

## Application - Ratios

From vertex $C$ of the right angle of $\triangle ABC,$ height $CK$ is dropped, and in $\triangle ACK$ bisector $CE$ is drawn. The line passing through point $B$ parallel to $CE$ meets $CK$ at point $F$. Prove that line $EF$ divides segment $AC$ in halves.

Since $\angle BCE=90° - \frac{\angle B}{2}$, we have $\angle BCE = \angle BEC$ and therefore $BE=BC$. Hence,

$CF : KF = BE : BK = BC : BK, \quad AE : KE = CA : CK = BC : BK.$

Let line $EF$ intersect $AC$ at point $D$. By Menelaus' theorem $\frac{AD}{CD} \cdot \frac{CF}{KF} \cdot \frac{KE}{AE} =1$. Taking into account the fact that $CF : KF = AE : KE ,$ we get our required statement. $_\square$

The following problem is a good example to invoke this theorem. Please try it out!

## Application - Collinearity

Converse of Menelaus' Theorem: Suppose three points $D,E,F$ are on sides (or extension) $AB,BC,AC$ respectively, such that $1$ or $3$ of them are in the extensions of the sides. Then points $D,E,F$ are collinear if and only if$\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=1.$

This can be used to prove Pascal's Theorem, which states that

Given 6 points (which can be coincident) on the circumference of a circle labelled $A, C, E, B, F,$ and $D$ in that order around the circle, the intersections of $AB$ and $DE$, $AF$ and $CD$, and $BC$ and $EF$ are collinear.

Let $G$ be the intersection of $\overline{CD}$ and $\overline{FA},$ let $H$ be the intersection of $\overline{AB}$ and $\overline{DE},$ and let $I$ be the intersection of $\overline{BC}$ and $\overline{EF}.$ We will prove these three points are collinear.

Let $U$ be the intersection of $\overline{CD}$ and $\overline{EF},$ let $V$ be the intersection of $\overline{AB}$ and $\overline{EF},$ and let $W$ be the intersection of $\overline{AB}$ and $\overline{CD}.$ By Menelaus in $\triangle UVW$ and line $HDE$, we have

$\dfrac {VH}{WH} \cdot \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} = 1.$

By Menelaus in $\triangle UVW$ and line $AGF$, we have

$\dfrac {VA}{WA} \cdot \dfrac {WG}{UG} \cdot \dfrac {UF}{VF} = 1.$

By Menelaus in $\triangle UVW$ and line $BCI$, we have

$\dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {UI}{VI} = 1.$

Multiplying these out, we get

$\dfrac {VH}{WH} \cdot \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {WG}{UG} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {UI}{VI} = 1.$

Upon rearranging, we get

$\dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {VH}{WH} \cdot \dfrac {WG}{UG} \cdot \dfrac {UI}{VI} = 1.$

Notice that by power of a point, we have

$\begin{aligned} \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} &= \dfrac {WD \times WC}{WA \times WB} \cdot \dfrac {VA \times VB}{VE \times VF} \cdot \dfrac {UE \times UF}{UC \times UD}\\\\ &= 1. \end{aligned}$

Thus, the product above simplifies as

$\dfrac {VH}{WH} \cdot \dfrac {WG}{UG} \cdot \dfrac {UI}{VI} = 1.$

So, by Menelaus, $G, H,$ and $I$ are collinear. $_\square$

## See Also

**Cite as:**Menelaus' Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/menelaus-theorem/