# 4 Variations on Ceva

The foundations of the rest of this essay are mainly composed of one result: Ceva’s theorem. Although it initially seems totally unrelated to Pappus’ theorem, our main result will eventually be a surprising and deep connection between the two.

In many ways, Ceva’s theorem is a much simpler result than Pappus’. The only real subtlety is becoming familiar with signed lengths. Let $$A$$ and $$B$$ be two points on an oriented line. The signed length from $$A$$ to $$B$$ is given by

\begin{align*} |AB| = \begin{cases} \hphantom{-}\|AB\| & \text{if the orientation points from A to B} \\ -\|AB\| & \text{otherwise}\end{cases} \end{align*}

where $$\|AB\|$$ denotes the distance between $$A$$ and $$B$$.

Theorem. (Ceva’s theorem)

Let $$ABC$$ be an oriented triangle in $$\mathbb{RP}^{2}$$ and let $$D$$ be a point (called the Ceva point) that doesn’t lie on the lines $$AB$$, $$BC$$ or $$AC$$. Let the lines $$AD$$, $$BD$$ and $$CD$$ meet the opposite sides of the triangle at $$Y, Z$$ and $$X$$ respectively (these points are called the intersection points). Then the signed lengths of segments satisfy the relation,

\begin{align*} \frac{|AX|}{|XB|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|} = 1. \end{align*}

Note: In this figure (and subsequent ones) we colour Ceva points green and intersection points orange. Also, the above figure demonstrates that $$D$$ need not be within the interior of $$ABC$$ (easier to see if you zoom out).

Proof. OK, cards on the table, we’re not actually going to see a full proof of Ceva’s theorem. It’s not a hard theorem to prove, but the simplest arguments I know all use the notion of a projective transformation/invariant and that’s beyond the scope of this essay. A good reference for such proofs is Section 15.4 in . However, proving the theorem’s Euclidean counterpart (i.e. we assume that $$D$$ is not a point at infinity) is pretty easy and gives a good intuition for why the result holds.6 So we will go over that proof.

If $$D$$ is contained within the interior of $$ABC$$ then all the terms in the left-hand-side of the desired equation are positive. On the other hand, if $$D$$ is not contained within the interior of $$ABC$$ then exactly two of the terms are negative. Therefore $$\frac{|AX|}{|XB|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|}$$ is always positive and we only have to show that

\begin{align*} \frac{\|AX\|}{\|XB\|} \cdot \frac{\|BY\|}{\|YC\|} \cdot \frac{\|CZ\|}{\|ZA\|} = 1. \end{align*}

The area formula for a triangle (area = $$\frac{1}{2}\times$$ height $$\times$$ base) tells us that, for any given height, a triangle’s area is proportional to the length of its base. We therefore have \begin{align*} \frac{\|BY\|}{\|YC\|}=\frac{\Delta(B,A,Y)}{\Delta(C,A,Y)}=\frac{\Delta(B,D,Y)}{\Delta(C,D,Y)} \end{align*} where $$\Delta(B,D,Y)$$ denotes the area of $$BDY$$. This implies \begin{align*} \frac{\|BY\|}{\|YC\|}=\frac{\Delta(B,A,Y)-\Delta(B,D,Y)}{\Delta(C,A,Y)-\Delta(C,D,Y)} = \frac{\Delta(B,A,D)}{\Delta(C,A,D)}. \end{align*} Similarly we get \begin{align*} \frac{\|AX\|}{\|XB\|}=\frac{\Delta(C,A,D)}{\Delta(C,B,D)} \quad \text{and} \quad \frac{\|CZ\|}{\|ZA\|}=\frac{\Delta(C,B,D)}{\Delta(B,A,D)}. \end{align*} Combining these three equalities gives \begin{align*} \frac{\|AX\|}{\|XB\|} \cdot \frac{\|BY\|}{\|YC\|} \cdot \frac{\|CZ\|}{\|ZA\|} = \frac{\Delta(B,A,D)}{\Delta(C,A,D)} \cdot \frac{\Delta(C,A,D)}{\Delta(C,B,D)} \cdot \frac{\Delta(C,B,D)}{\Delta(B,A,D)} = 1. \end{align*}

$$\square \qquad$$

One of the coolest things about Ceva’s therorem is that you can glue multiple copies of the theorem to itself. What does that mean? Let’s take a look at the following set up.

Note that, although the first Ceva point $$D$$ may be chosen freely, the second Ceva point $$D'$$ is constrained to ensure that the two Ceva configurations share the same intersection point along the line $$(BC)$$. When two Ceva configurations share the intersection point along an edge like this, we say that they are compatible along that edge.

Applying Ceva’s theorem to the left-hand triangle tells us that $$\frac{|AX|}{|XB|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|} = 1$$, while applying it to the right-hand triangle tells us that $$\frac{|A'Z'|}{|Z'C|} \cdot \frac{|CY|}{|YB|} \cdot \frac{|BX'|}{|X'A'|} =1$$. We note that requiring the quadrilateral $$ABA'C$$ be oriented forces the induced orientations on $$ABC$$ and $$A'BC$$ to disagree along the shared line $$BC$$; this is why the $$\frac{|BY|}{|YC|}$$ and $$\frac{|CY|}{|YB|}$$ terms are inversed. Therefore, by multiplying these two equalities, we get a result that only depends on points that lie on the perimeter of the quadrilateral $$ABA'C$$, namely \begin{align*} \frac{|AX|}{|XB|} \cdot \frac{|BX'|}{|X'A'|} \cdot \frac{|A'Z'|}{|Z'C|}\cdot \frac{|CZ|}{|ZA|} = 1. \end{align*}

But there’s no reason to stop with just two copies glued together. We can glue as many as we like! That said, the results won’t get much more interesting: we’ll just prove that an ever-growing product of signed lengths along the perimeter evaluates to one. No, things get really interesting when we apply this perspective to triangulated surfaces.7

Let’s consider a triangulated sphere, i.e. a sphere whose surface has been divided up into triangles. Now let’s imagine that, one by one, we fill each of the triangles with compatible Ceva configurations until exactly one triangle remains unfilled. What would that look like? Here’s a sphere I prepared earlier.

You can move the Ceva point labelled $$D$$ to change the initial Ceva configuration and see how the others adapt (although moving the point into the exterior of its containing triangle will break things as drawing on a 3D surface is hard). Try to find the triangle without an interior Ceva point. To get a good view of it, I’d recommend using the full-screen button.

If you’ve found the empty triangle, you’ll have seen that its corners are labelled $$A,B,C$$ and the intersection points caused by the surrounding Ceva configurations are labelled $$X,Y,Z$$. Exactly as we described before, applying Ceva’s theorem to each triangle, multiplying all the resulting equalities together and then cancelling all possible terms, leads to an equality that only depends on points that lie on the perimeter i.e. our one remaining triangle. And what is that equality? \begin{align*} \frac{|AX|}{|XB|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|} = 1. \end{align*} Huh, that’s funny. It’s exactly the conclusion we’d draw if $$X,Y$$ and $$Z$$ were defined by a Ceva configuration on $$ABC$$. If only Ceva’s Theorem had some kind of converse…

Theorem. (Converse to Ceva’s theorem)

Let $$ABC$$ be a triangle in $$\mathbb{RP}^{2}$$, let $$X,Y$$ and $$Z$$ be points that lie on $$AB,BC$$ and $$CA$$ respectively, such that \begin{align*} \frac{|AX|}{|XB|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|} = 1 \end{align*} Then $$AY , \ BZ$$ and $$CX$$ intersect in a single point.

Proof. Let $$D = AY \cap BZ$$ and $$\tilde{X}=CD \cap AB$$. Then, by Ceva’s theorem,

\begin{align*} \frac{|A\tilde{X}|}{|\tilde{X}B|} \cdot \frac{|BY|}{|YC|} \cdot \frac{|CZ|}{|ZA|} = 1 \text{, and therefore } \frac{|A\tilde{X}|}{|\tilde{X}B|} = \frac{|AX|}{|XB|} \end{align*}

which, as $$X$$ and $$\tilde{X}$$ both lie on $$AB$$, implies $$X = \tilde{X}$$. $$\square$$

So, returning to our triangulated sphere, the equality we derived on the empty triangle exactly tells us that $$AY , \ BZ$$ and $$CX$$ intersect in a single point.

I find this really beautiful. Just by knowing Ceva’s theorem and how to glue copies of it together, we can automatically know the incidence result that $$AY,BZ$$ and $$CX$$ intersect in a single point. An important point to realise is that no part of the argument we just made uses the fact that our surface was a sphere. We therefore get the following, more general (if also a bit more wordy), statement.

Proposition. (Compatible Ceva configurations)

Let $$\mathcal{S}$$ be a (compact, orientable) surface, let $$\mathcal{T}$$ be a triangulation of $$S$$8 and let $$T$$ be a distinguished triangle in $$\mathcal{T}$$. Suppose that every triangle in $$\mathcal{T}$$ other that $$T$$ is equipped with a Ceva configuration such that adjacent configurations are compatible along their shared edge. Then $$T$$ admits a Ceva configuration that is compatible with all its adjacent configurations.

Now we just have to prove Pappus’ theorem using this result.

1. You can get a pretty good picture of what the theorem looks like when $$D$$ is a point at infinity by moving $$D$$ very far away from $$ABC$$ in the interactive figure (zooming out, and then back in with the mouse scroll helps a lot for this).↩︎

2. Or, to be more precise, compact orientable triangulated 2-manifolds.↩︎

3. In other words, $$\mathcal{T}$$ is a collection of non-overlapping triangles that cover all of $$S$$.↩︎