Discovering Ceva's Theorem

This result deals with line segments that go from one vertex of a triangle to a point on the opposite side. In tribute to the Italian mathematician Giovanni Ceva, these are sometimes called cevians of the triangle. In the figure below, AW, AX, BY, and CZ (in red) are all examples of cevians for the triangle ABC (in blue).

Although the term cevian may be new to you, the concept is certainly not. You have seen examples. For instance, the medians, the altitudes, and the angle bisectors are examples of cevians.

You have probably noticed that, in each of these examples, the three cevians all go through a single point. Ceva's Theorem gives a condition that determines whether or not three cevians from the three vertices of a triangle will have this concurrency property.

Open the next sketch (See file ex3_1.gsp). This shows a generic triangle along with three cevians, one from each vertex. We have also included the ratios into which the endpoints of the cevians divide their corresponding sides of the triangle. Use this sketch to perform the following experiments. In each case, record the values for the ratios CX/XB, AY/YC, and BZ/ZA.

1. Set X and Y to be as close as you can make them to the midpoints of BC and CA respectively. This means that the corresponding ratios should be as close as you can make them to 1.000. Now slide Z until CZ goes through the intersection point of AX and BY. In other words, until the three cevians are concurrent. What is the value of the final ratio BZ/ZA?

    

2. Move X and Y so that they are each about 1/3 of the way along their segments. That means the ratios will be as close as you can get them to 0.500. Now slide Z until the cevians are concurrent. What is the value of BZ/ZA?

    

3. Move X so that it is as close as you can get to the midpoint of CB. This means that the ratio CY/YB should be as close as you can get to 1.000.

    

   1. Move Y so that the ratio AY/YC is as close as you can get to 2.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   2. Move Y so that the ratio AY/YC is as close as you can get to 3.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   3. Move Y so that the ratio AY/YC is as close as you can get to 4.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   Can you see any pattern here?

    

4. Select the three measurements and then use Calculate... from the Measure menu to compute the product of these three ratios. Return to the sketch and move the vertices and the cevians into different positions, but make sure that the cevians are concurrent. What is the value of the product of the three ratios in each case?

    

By this stage you should have discovered:

Ceva's Theorem:

If the three cevians AX, BY and CZ are concurrent, then the product of the ratios (CX/XB) (BZ/ZA) (AY/YC) = 1.

 

The ratios are taken in the same "order," that is, either all counterclockwise (as in the statement of the theorem above) or, equivalently, all clockwise. In the clockwise case the condition would be

(BX/XC) (CY/YA) (AZ/ZB) = 1.

The Extended Version of Ceva's Theorem

First we should consider what we mean by the ratio in which a point divides a segment if the point is not actually inside the segment. This may sound kind of silly, but consider the following diagram.

where X is on the line determined by A and B but is not between them. Although you might well say that X does not divide the segment at all, you can still make perfectly good sense of the ratio AX/XB. In the picture below, the red segment is equal in length to AX and the blue on to XB

.

You can open a Sketchpad version of this ( See file ex3_2.gsp). Once you have done that, drag the point X along the line and observe the changes in the ratio.

Now we can create a sketch that will check Ceva's Theorem for more general cevians. Open a New Sketch. In this sketch:

1. draw three blue lines to form a triangle. Label the vertices of this triangle A, B, and C,
2. construct points X (on BC) and Y (on AC) then construct two dashed red lines AX and BY,
3. construct the intersection point, O, of these two lines,
4. construct the dashed red line CO and construct its intersection point, Z, with the line AB,
5. construct the thick green segments CX, XB, BZ, ZA, AY, and YC and use them to measure the ratios CX/XB, BZ/ZA, and AY/YC,
6. select these ratios and use them to calculate the product (CX/XB) (BZ/ZA) (AY/YC).

Now you can drag X and Y along their respective lines and the three cevians should remain concurrent. What happens to the product of the ratios?

In other words, Ceva's Theorem is still true when the cevians do not all lie in the interior of the triangle.

Proving Ceva's Theorem

To prove Ceva's Theorem in the case where all the cevians lie in the interior of the triangle, we have to show that if the Cevians are concurrent then the product of the ratios is 1. Use the point of concurrency, O, to form the three triangles (colored red, green, and blue in the figure below) AOC, COB, and BOA.

From your homework for Lesson 2, you know, for instance, that the ratio Area(COB)/Area(AOC) is equal to the ratio BZ/ZA. Using this, and similar equalities, you can complete the proof. This is one of your homework exercises.

Homework

Complete the exercises in Lesson 3 homework and submit them to the MTL handin system.