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High School Geometry: Help and Review13 chapters | 162 lessons

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.

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.

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*|

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

Take a look at the stained glass window again.

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.

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

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.

Now, suppose that we are trying to figure out possible lengths for *NT*, *OR*, and *MS*. Consider the following two possibilities:

- |
*NT*| = 2, |*OR*| = 4, |*MS*| = 8 - |
*NT*| = 14, |*OR*| = 3, |*MS*| = 6

Which of these sets of lengths could actually be the lengths of *NT*, *OR*, and *MS* based on what we just found using Ceva's theorem? We know that |*NT*| × |*OR*| × |*MS*| must equal 168. If we check the first set of lengths, we get

- |
*NT*| × |*OR*| × |*MS*| = 2 × 4 × 8 × = 64

This is not equal to 168, so this is not a possibility. The second set gives the following:

- |
*NT*| × |*OR*| × |*MS*| = 14 × 3 × 6 = 168

Ding, ding, ding! We have a winner. It could be the case that |*NT*| = 14, |*OR*| = 3, and |*MS*| = 6, since this would result in their product equaling 168. It would take a bit more to figure out if these are the actual side lengths, but it does give a possibility to try to work with. Wow! There's so much that can be found out about a triangle using Ceva's theorem!

**Ceva's theorem** 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*|

This theorem can be used to analyze many different parts of a triangle. Because triangles occur extremely often in the world around us, it is a very useful theorem for real-world applications in art, construction, physics, and astronomy, as well as in purely mathematical applications.

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High School Geometry: Help and Review13 chapters | 162 lessons

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