Cevians

Van Aubel's Theorem

As shown in the diagram, $ABC$ is a triangle with concurrent cevians $CD, BE, AF$. Then, van Aubel's Theorem states that $\frac{CE}{EA} + \frac{CF}{FB} = \frac{CG}{DG}$.

Proof. Applying Menalaus Theorem on $\triangle ADC$ and $\triangle CDB$ respectively gives :

$\frac{CE}{EA} \cdot \frac{AB}{BD} \cdot \frac{DG}{GC} = 1$ and $\frac{CF}{FB} \cdot \frac{AB}{AD} \cdot \frac{DG}{GC} = 1$

So, $\frac{CE}{EA} = \frac{BD}{AB} \cdot \frac{CG}{DG}$, and $\frac{CF}{FB} = \frac{AD}{AB} \cdot \frac{CG}{GD}$. Adding these two equalities gives the desired conclusion.

We can generalize this theorem for points outside the triangle by using signed lengths. The proof is very similar.

Again we apply Menelaus theorem on $\triangle ADC$ and $\triangle CDB$. This now gives us $\frac{CE}{EA}\cdot\frac{AB}{BD}\cdot\frac{DG}{GC}=-1$ and $\frac{CF}{FB}\cdot\frac{BA}{AD}\cdot\frac{DG}{GC}=-1$.

So, $\frac{CE}{EA}=-\frac{BD}{AB}\cdot\frac{GC}{DG}$ and $\frac{CF}{FB}=-\frac{AD}{BA}\cdot\frac{GC}{DG}$.

Adding these two equalities gives $\frac{CE}{EA}+\frac{CF}{FB}=-\frac{BD}{AB}\cdot\frac{GC}{DG}-\frac{AD}{BA}\cdot\frac{GC}{DG}$.

We simplify this result using the rules $XY = -YX$ and $XY+YZ = XZ$, when $X$, $Y$ and $Z$ are collinear.

$\frac{CE}{EA}+\frac{CF}{FB} = -\frac{BD}{AB}\cdot\frac{GC}{DG}-\frac{AD}{BA}\cdot\frac{GC}{DG} = \frac{DB}{AB}\cdot\frac{GC}{DG}+\frac{AD}{AB}\cdot\frac{GC}{DG} = \frac{AB}{AB}\cdot\frac{GC}{DG} = \frac{GC}{DG} = \frac{CG}{GD}$.

We can now conclude $\frac{CE}{EA}+\frac{CF}{FB} = \frac{CG}{GD}$.

Cevians

Cevians are defined as segments drawn from a vertex of a triangle to its opposite side. For example, angle bisectors, medians, and altitudes are all cevians:

In geometry, we are often interested in cevians that are concurrent, as they provide many interesting results like Ceva's Theorem and Menelaus' Theorem. In fact, the angle bisectors of a triangle are all concurrent, and the same goes for the medians and altitudes.