(by Polymath) This page is part of the Advanced High School Math Project.

For most of the Geometry proofs on this site, I hope to show proofs that are considerably easier than those on most other sites. This one is an exception, though. This proof is just spiffed up version of the one on this site, which is already pretty good. But any series of posts on advanced geometry absolutely has to include a proof of Ceva's theorem—it's just too important to leave out. So I'm going to include this proof here.

If you draw a segment from one vertex of a triangle to a point on the opposite side*, that segment is called a cevian (pronounced "chavian" to rhyme with "avian"). They are named after Giovanni Ceva, a 17th century mathematician, who studied them and, obviously, came up with the theorem we're proving. If you draw a cevian from each vertex of a triangle, of course, chances are pretty slim that all three will *concur*, which a fancy way of saying that they all run through the same point. The point of Ceva's theorem is that it provides a relatively simple calculation (using ratios) you can do that will tell you whether or not the three cevians really do concur. The theorem actually works in two directions: if the calculation holds then the segments concur, and if the segments concur then the calculation holds. We'll prove the second one first, right after two simple preparatory facts.

The first fact is about ratios. If you know two ratios are equal (we'll call them *k* in the first equation below), then subtracting their numerators and subtracting their denominators gives you a new ratio that's also equal to the first two.

We start with

which easily turns into

and then the ratio we're constructing by subtracting clearly also equals *k*.

We'll see how that's useful below.

The other fact is a geometric one about area. It is well-known, but important to the proof, so I'll quickly state it here. It says that when a cevian divides a triangle, the areas of the two resulting triangles stand in the same ratio as the two parts of the side the cevian is drawn to. In other words, in the diagram

the ratio (x:y) equals the ratio (area of ABT:area of CBT). This is because they have the same height (though it falls inside one triangle and outside the other); simply using the formula for the area of the triangles makes it clear why the ratio of the areas is equal to (x:y).

Once you know those two facts, Ceva's theorem isn't all that hard to prove. The basic diagram is below.

We're supposing that the three cevians intersect at a single point P. Note that the area theorem from above applies twice here for the ratio BX:XC. It applies to the light-colored triangles BXP and CXP, and it applies to the combined blue and combined green triangles BXA and CXA. In other words,

This is exactly the ratio situation discussed in the first of our preliminary facts. The two area ratios are the ones we'll subtract, and we showed above that the result must be equal to the original ratio, so that

By exactly the same reasoning, we get

And finally, we put those last three ratios together into one calculation. You'll notice that each triangle's area appears once in a numerator and once in a denominator. So we get a perfect result if we multiply those ratios:

This is the main equation for Ceva's theorem. One way to remember it is to think about walking around the triangle. Pick any vertex to start (I chose B), and walk around in either direction (I chose counter-clockwise). Insert each segment into the ratios as you come to it, and then they'll be in the right place.

As a matter of logic, it isn't true that a theorem implies its own converse. Sometimes, though, you can use a theorem to prove its own converse, and this is one of those times. That is, now that we know that concurrent cevians imply that the above product equals 1, we can prove that the product equaling 1 also implies that the cevians concur.

Suppose you are given segments BY and CZ in the picture above. You're told that a cevian from A lands on segment BC at point N, and that the relevant product using that cevian equals 1, even though you don't know whether that cevian passes through P where the other two intersect. That is, suppose

The cevian that *does* go through P clearly lands at X, and the first part of the theorem therefore says that the product for point AX also equals 1. Comparing the two equations makes it obvious that BX/XC and BN/NC must be the same ratio. Since only one point on segment BC can create that ratio, we have to conclude that X and N are the same point, which means our hypothetical segment AN must really be AX, which goes through P. And that's what we wanted to prove.

Thus, Ceva's theorem states that three cevians of a triangle concur if and only if the ratios of the segment lengths they create satisfy the main equation above.

*If you allow the idea of *directed* segment lengths (positive or negative depending on their direction), then the point on the opposite side where a cevian lands need not be between the endpoints of the opposite side—it could fall outside the triangle. But that possibility complicates the explanation, so we won't deal with that here.