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Ceva's Theorem: Applications & Examples

Instructor: Laura Pennington

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

Ceva's theorem is an interesting theorem that has to do with triangles and their various parts. This lesson will state the theorem and discuss its application in both real-world and mathematical examples.

Ceva's Theorem

An artist has created a triangular stained glass window and has one strip of siding left before completing the window. She needs to figure out the length of the last side based on the lengths of the other sides, as shown in the image.


Ceva1


We need to know how the line segments relate, and then use that relationship to find the length of FB (the missing side length). It just so happens we have a nice theorem that will help this artist to understand exactly how these parts of the triangular window relate and find the length of the last strip of siding.

Before we give the theorem, let's review a definition that will aid in understanding the theorem. A cevian of a triangle is a line segment that runs from any of the vertices of the triangle to the side opposite that vertex. In terms of the triangular window, the cevians of the triangle are the line segments AD, BE, and CF, because each of these runs from a vertex of the triangle to a point on the side opposite that vertex.

Okay, now we can explore this theorem that is going to help the artist find what she's looking for! The name of the theorem is Ceva's theorem, and it states that if we have a triangle ABC and points D, E, and F are on the sides of the triangle, then the cevians AD, BE, and CF intersect at a single point if and only if

  • |BD| × |CE| × |AF| = |DC| × |EA| × |FB|


Ceva2


Now that we know the theorem, let's apply it to the artist's scenario.

Applying Ceva's Theorem

Take a look at the stained glass window again.


Ceva1


Notice that the line segments AD, BE, and CF within the window are cevians of the triangular window. Also, notice that those cevians meet at a single point. Are you starting to see how Ceva's theorem might be applied to this situation?

Since the cevians of the triangular window intersect at exactly one point, it follows by Ceva's theorem that |BD| × |CE| × |AF| = |DC| × |EA| × |FB|. Ah-ha! These are the sides that make up the siding of the triangular window. Furthermore, we have the following lengths for some of those sides.

  • |AF| = 5 inches
  • |BD| = 3 inches
  • |DC| = 10 inches
  • |CE| = 4 inches
  • |EA| = 3 inches

Thus, we can plug these into this relationship to get the following:

  • 3 × 4 × 5 = 10 × 3 × |FB|

The only unknown is |FB|, and that is the side length that the artist was looking for! All we have to do is simplify this equation and solve for |FB|.

  • 60 = 30 × |FB|

Divide both sides of this equation by 30 to get the following:

  • 2 = |FB|

There we have it! By Ceva's theorem, it must be the case that the last strip of siding will have length 2 inches.

This is all pretty fascinating! Let's take a look at another example.

Another Example

The image shows a triangle RST and random points M, N, and O, on the sides of the triangle.


Ceva3


Suppose that |SN| = 4 units, |TO| = 7 units, |RM| = 6 units. Now, suppose that we draw in the cevians RN, SO, and TM and they intersect at exactly one point. Based on this, what must be true about NT, OR, and MS?

Since the cevians intersect at exactly one point, by Ceva's theorem, we have that |SN| × |TO| × |RM| = |NT| × |OR| × |MS|. We also have lengths for SN, TO, and RM, so we have the following:

  • |SN| × |TO| × |RM| = 4 × 7 × 6 = 168 = |NT| × |OR| × |MS|

Therefore, it must be true that the product of the lengths of NT, OR, and MS is 168.

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