# Menelaus' Theorem

**Menelaus' Theorem** relates ratios obtained by a line cutting the sides of a triangle. The converse of the theorem 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' theorem states that if a line intersects \(\triangle ABC\) or extended sides at points \(D\), \(E\) and \(F\), the following statement holds: \[\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=1\]

**Proof:**

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, \[\frac{AD}{DB}\times\frac{BE}{EC}\times\frac{CF}{FA}=\frac{AA'}{BB'}\times\frac{BB'}{CC'}\times\frac{CC'}{AA'}=1\]

**Note:**

(1) 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).

(2) 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 \(AD\times BE\times CF=DB\times EC\times FA\).

## Variations

## Application-Ratios

https://brilliant.org/problems/yellow-quaditerial/

## From vertex \(C\) of the right angle of \( \Delta ABC\) height \(CK\) is dropped and in \(\Delta ACK\) bisector \(CE\) is drawn. Line that passes 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, \[\begin{align} CF : KF = BE : BK = BC : BK & AE : KE = CA : CK = BC : BK \end{align}\]

Let the line \(EF\) intersect \(AC\) at point \(D\) . By Menelaus's Theorem \(\dfrac{AD}{CD} . \dfrac{CF}{KF} . \dfrac{KE}{AE} =1\) . Taking into account that \(CF : KF = AE : KE \) we get our required statement .

## Application-Collinearity

**Cite as:**Menelaus' Theorem.

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