Ceva’s Theorem

Ceva’s Theorem and Converse of 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. 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,

\[\frac{AF}{FB}.\frac{BD}{DC}.\frac{CE}{EA} = 1\]

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,

\[\frac{AF}{FB}.\frac{BD}{DC}.\frac{CE}{EA} = 1\]

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,

\[\frac{AF}{FB}=\frac{[AFB]}{[FBP]}=\frac{[AFC]}{[FBC]}\]

When you subtract the triangle area of the second quality from the area of the first equality, you get,

\[\frac{AF}{FB}=\frac{[APC]}{[BPC]}\]

Similarly, 

\[\frac{BD}{DC}=\frac{[APB]}{[APC]} and \frac{CE}{EA}=\frac{[BPC]}{[APB]}\]

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

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. 

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.


  1. 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/AFAB/AC and CD/CE = CA/BC

When you multiply all these three equations, you would get,

BF/BD . AE/AF . CD/CEBC/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.


FAQ (Frequently Asked Questions)

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, as shown in the figure below.


When you use the signed lengths of the segments, you get,

BD/DCABcosABC/ACcosBCA,

CE/EA-BCcosBCA/-ABCcosBACBCcosBCA/ABCcosBAC, and,

AFFB-ACcosBAC/-BCcosABC = ACcosBAC/BCcosABC

Simplifying this gives you

BD/DC CE/EA AF/FBABcosABC/ACcosBCA BCcosBCA/ABCcosBAC ACcosBAC/BCcosABC = 1

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.