Introduction to Ceva’s Theorem
The Ceva’s theorem is a theorem on the triangles that lie in the Euclidean plane geometry. It considers the ratio of the side lengths of the triangle that are divided by cevians.
Take an example of a triangle DEF, it’s lines DO, EO and FO be sketched over its vertices to a common point that is O, so that it could meet the opposite sides at G,H and I.
Ceva's theorem is a theorem of affine geometry, in the context that it may be stated and proved without the use of the concepts of angles, areas, and lengths (except for the ratio of the lengths of two given line segments which are collinear). Therefore, it is true for triangles in any affine plane over any field. The Ceva’s theorem is helpful in proving the concurrence of cevians in the triangles and is commonly used in the Olympiad geometry. In this article, we will learn about Ceva's Theorem and the converse of Ceva’s Theorem in detail.
Statement of Ceva’s Theorem
Consider a triangle ABC with a point P that lies inside the triangle. Consider the lines AP, BP, and CP to hit the BC, CA, and AB at the points D, E, F, D, E, and F, respectively.
According to Ceva’s theorem,
AF/FB . BD/DC . CE/EA = 1
The converse of Ceva’s theorem is true as well. if the points D, E and F lie on the sides BC, CA, and AB, respectively, in a way that,
AF/FB . BD/DC . CE/EA = 1
then the lines AD, BE and CF are concurrent at the point P.
Ceva’s Theorem Proof
Consider the same triangle.
Since the triangles △AFP and △FBP possess the same altitudes, you can write,
AF/FB = [AFB]/[FBP] = [AFC]/[FBC]
When you subtract the triangle area of the second quality from the area of the first equality, you get,
AF/FB = [APC]/[BPC]
Similarly,
BD/DC = [APB]/[APC] and CE/EA = [BPC]/[APB]
When you multiply the previous three equations together, you would get,
AF/FB . BD/DC . CE/EA = APC/BPC . APB/APC . BPC/APB
Converse of Ceva’s Theorem
There are multiple proofs of the theorem that have been made. Two amongst the other are discussed below:
The first one is shown just by using the basic properties of the areas of a triangle. However, it is not that easy because many cases must be considered, that depends on the position that point O has.
Another proof is shown using vectors and barycentric coordinates and it seems more natural and dependent on the case. It works on any field in any affine plane.
First Approach
The left-hand side’s sign is positive for two reasons. When O is inside the triangle, in this case, all three ratios seem positive. When O is outside the triangle, in that case, one is positive and the remaining two seem negative.
If you want to check the magnitude, the area of the triangle with a given height is proportional to the base of the triangle.
Consider the following triangle.
As you know that according to Ceva’s theorem,
AF/FB . BD/DC . CE/EA = 1
You have to prove that the Cevians AD, BE and CF are concurrent at a point.
Let AD and BE meet at an arbitrary point P. Let the third Cevian through the point P be CK. According to the previous proof, you have
(BD/CD) . (CE/AE) . (AK/BK) = 1
But you have assumed that
(BD/CD) . (CE/AE) . (AF/BF) = 1
Hence, you can say that
(AK/BK) = (AF/BF)
Therefore, it is obvious that K = F and you can conclude that AD, BE and CF are concurrent.
It was Giovanni Ceva, who published this theorem in his work, De lineis rectis in 1678. But there is evidence that it was proved much earlier by the eleventh-century king of Zaragoza named Yusuf Al-Mu’taman ibn Hud.
Second Approach
The three points D,E,F which are not collinear along with a point O, which are on the same plane, Gets three unique numbers with respect to barycentric coordinates of O. These unique numbers when added together equals 1.
In Ceva’s theorem, the point O should not belong to any line that passes through two vertices of the triangle. This shows that those three unique numbers are not equal to 0.
After rearranging the last equation by replacing X from F , the new equation will have a vector on its left-hand side, with the same direction as the line EF. On its right-hand side, it will have the same direction as that of line DE.
As D,E and F aren’t collinear, the directions of these lines will be different. It Shows the two members of the equation are equal to the zero vector and it also shows that the signed ratio of the length which are collinear line segments are on the left-hand side of the fraction.
Solved Examples
1. Prove that if X, Y, and Z are the midpoints of the sides, then the three cevians are concurrent.
Solution: If you consider the same triangle,
then you can say that D, E, and F are the midpoints of their respective sides BC, AC, and AB. Therefore, you can say that AE = EC, CD = DB and BF = FA.
This satisfies the Ceva’s theorem that says that
AF/FB . BD/DC . CE/EA = 1
Hence, the intersection occurs at the centroid, which proves that the three cevians are concurrent.
2. Prove that the cevians perpendicular to the opposite sides are concurrent.
Solution: Consider D, E, and F to be the feet of the altitudes.
You must know that BFC∼△BDA.
Similarly,
△AEB∼△AFC, and △CDA∼△CEB.
Hence, you can say that
BF/BD = BC/BA , AE/AF = AB/AC and CD/CE = CA/BC
When you multiply all these three equations, you would get,
BF/BD . AE/AF . CD/CE = BC/BA . AB/AC . CA/BC
If you rearrange the left-hand side, you would get,
AF/FB . BD/DC . CE/EA = 1
Therefore, you can say that the three altitudes are coinciding at one single point, which is the orthocenter.
FAQs on Ceva’s Theorem
1. How to explain Ceva’s theorem?
Ceva’s theorem is a theorem on triangles in plane geometry. If a triangle ABC is given, let the lines AO, BO and CO be drawn from the vertices of the triangle to a common point O, which is not on one of the sides of the triangle ABC. Let these sides meet to the opposite sides at D, E and F respectively. The segments AD, BE, and CF are called Cevain. Then, by using the signed lengths of these segments, you would get,
AF/FB . BD/DC . CE/EA = 1
2. How to use Ceva’s theorem to prove that all three altitudes of a triangle are concurrent at one point?
Using Ceva’s theorem, you have to prove that all three altitudes of a triangle are concurrent at one point.
The converse of Ceva’s theorem states that if the product of the ratios of the three sides of a triangle when divided by three points results and is equal to 1, then it means that the lines that join these three points to the opposite vertices of the triangle are concurrent in nature.
Consider a triangle ABC to be any arbitrary triangle and let AD, BE and CF be the altitudes of the triangle.
When you use the signed lengths of the segments, you get,
BD/DC = ABc ABC/ACcosCA,
CE/EA = - BCcosBCA/ - ABCcosBAC = BCcosBCA/ABC sBAC, and,
AFFB = - ACcosBAC/ - BCcosABC = ACcosBAC/BCcocABC
Simplifying this gives you
BD/DC CE/EA AF/FB = ABcosABC/ACcosBCABCcosBCA/ABCcosBAC ACcosBAC/BCcosAB
This shows that the product of the ratios of the divisions of the three sides of the triangle ABC by the three points D, E and F equals 1.
By the converse of Ceva’s theorem, you can say that the lines AD, BE and CF are concurrent.
However, AD, BE and CF are the altitudes of the arbitrary triangle.
Therefore, you can say that all the three altitudes of the triangle ABC are concurrent.
3. How to generalise Ceva’s theorem?
Ceva’s theorem can be generalised into simplexes that are higher dimensional by using barycentric coordinates. An n-simplex cevian can be defined as a ray from each vertex upto a point which is on the opposite (n-1) face. The cevians look concurrent only if to the vertices, a mass distribution can be assigned so that each cevian at its centre of mass, intersects the opposite facet. The cevian’s intersection point is the simplex’s centre of mass. It was routh’s theorem that helped in knowing the area of the triangle which is made by the three cevians if they are not concurrent. If you set the area equal to zero and then solve it, Ceva’s theorem can be made.
4. What is Menelaus’s theorem?
Menelaus’s theorem is the same as Ceva’s theorem. The only difference between the two is their equation is different only by the sign. Menelaus of Alexandria, after whose name the Menelaus’s theorem got its name. The theorem is about triangles and their proposition in plane geometry. For example, we have a triangle names DEF, along with a transversal line that passes EF, DF and DE at the points G,H and I, where G,H and I are different from D,E and F. By using signed length of segments, it means to see that the length DE is either positive or negative as it depends on if D is in the left or right of E, in a fixed line.
5. What is the conclusion of Ceva’s theorem?
When you generalise the Ceva’s theorem to a higher-dimensional simplex that enhances the end result of the theorem that the product of some particular Ratios is equal to 1. If you start from a point in a simplex, a point is described on every k-face. This very point is the Cevian’s foot which is from the k-face, that is the opposite vertex, in a (k+1)- face, which consists of it, to a point that is already described on the (k+1)- face. Each point then divides the face, where it lies in the lobes. For more detail, search Vedantu NCERT Solutions for Class 12 Chemistry - Vedantu.
