Ceva's Theorem: Statement, Proof and Uses

Quick answer: Ceva's theorem says that in triangle $ABC$, with points $D$, $E$ and $F$ on sides $BC$, $CA$ and $AB$, the lines $AD$, $BE$ and $CF$ meet at one point if and only if $\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1$. It can be proved using ratios of areas of triangles. It gives quick proofs that the medians and the angle bisectors of a triangle are concurrent.

You may know that the three medians of a triangle meet at one point (the centroid), and that the three angle bisectors also meet at one point (the incentre). Each of these facts usually has its own proof.

In 1678, the Italian mathematician Giovanni Ceva published a single result that covers all such cases: a simple test, using the ratios in which three lines divide the sides, for whether the lines meet at a point.

Syllabus note: Ceva's theorem is not part of the NCERT Class 10 syllabus. It is enrichment material, useful for Olympiad preparation and for students who enjoy geometry. The proof below uses only the idea that triangles with the same height have areas in the ratio of their bases.

What You Will Learn

• What a cevian is
• The statement of Ceva's theorem and its converse
• A proof using ratios of areas
• How Ceva's theorem proves that the medians are concurrent
• How it proves that the angle bisectors are concurrent

1. What Is a Cevian?

Definition: A cevian is a line segment joining a vertex of a triangle to a point on the opposite side (or on the opposite side extended).

Medians, angle bisectors and altitudes are all cevians. Three lines are concurrent if they pass through the same point.

2. Statement of Ceva's Theorem

Ceva's theorem: Let $D$, $E$ and $F$ be points on sides $BC$, $CA$ and $AB$ of triangle $ABC$. The cevians $AD$, $BE$ and $CF$ are concurrent if and only if $\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1$

             A
            / \
           /   \
        F /  P  \ E
         /       \
        B----D----C
  AD, BE and CF meet at P

Memory aid: start at a vertex and go round the triangle in one direction: $A \rightarrow F \rightarrow B \rightarrow D \rightarrow C \rightarrow E \rightarrow A$.

3. Proof Using Ratios of Areas

Suppose $AD$, $BE$ and $CF$ meet at $P$. Write $[XYZ]$ for the area of triangle $XYZ$.

Step 1: Triangles with the same height. Triangles $ABD$ and $ACD$ have the same height from $A$, so their areas are in the ratio of their bases:

$\frac{[ABD]}{[ACD]} = \frac{BD}{DC}$

Similarly, triangles $PBD$ and $PCD$ have the same height from $P$:

$\frac{[PBD]}{[PCD]} = \frac{BD}{DC}$

Step 2: Subtract. If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a - c}{b - d} = \frac{a}{b}$ (when $b \ne d$). So:

$\frac{BD}{DC} = \frac{[ABD] - [PBD]}{[ACD] - [PCD]} = \frac{[PAB]}{[PCA]} \qquad \text{...(1)}$

Step 3: Repeat for the other two sides.

$\frac{CE}{EA} = \frac{[PBC]}{[PAB]} \qquad \text{...(2)}$

$\frac{AF}{FB} = \frac{[PCA]}{[PBC]} \qquad \text{...(3)}$

Step 4: Multiply (1), (2) and (3).

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{[PCA]}{[PBC]} \times \frac{[PAB]}{[PCA]} \times \frac{[PBC]}{[PAB]} = 1$

Hence proved. The converse is also true: if the product equals $1$, the three cevians are concurrent.

4. Application 1: The Medians Are Concurrent

For medians, $D$, $E$ and $F$ are the midpoints of the sides, so:

$\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = 1$

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1 \times 1 \times 1 = 1$

Conclusion: By the converse of Ceva's theorem, the three medians are concurrent. They meet at the centroid.

5. Application 2: The Angle Bisectors Are Concurrent

Let $a = BC$, $b = CA$ and $c = AB$. By the angle bisector theorem (the bisector of an angle divides the opposite side in the ratio of the other two sides):

$\frac{BD}{DC} = \frac{c}{b} \qquad \frac{CE}{EA} = \frac{a}{c} \qquad \frac{AF}{FB} = \frac{b}{a}$

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{b}{a} \times \frac{c}{b} \times \frac{a}{c} = 1$

Conclusion: The three angle bisectors are concurrent. They meet at the incentre.

Ceva's theorem can also be used to show that the three altitudes are concurrent (at the orthocentre), but that proof needs trigonometry.

6. Summary Table

7. Tips and Common Mistakes

Tip: Write the three ratios in order going round the triangle in one direction, then multiply.

Tip: To prove three cevians are concurrent, show the product of the ratios is $1$ and quote the converse of Ceva's theorem.

Common Mistake: Mixing up the order of the segments, for example writing $\frac{FB}{AF}$ instead of $\frac{AF}{FB}$. Keep going round in the same direction.

Frequently Asked Questions

What is Ceva's theorem?

In triangle ABC with D, E and F on BC, CA and AB, the lines AD, BE and CF meet at one point if and only if (AF/FB) × (BD/DC) × (CE/EA) = 1.

What is a cevian?

A line segment from a vertex of a triangle to a point on the opposite side.

How does Ceva's theorem prove that the medians are concurrent?

Each median cuts its side in the ratio 1 : 1, so the product of the three ratios is 1, and by the converse of Ceva's theorem the medians are concurrent.

Is Ceva's theorem in the Class 10 syllabus?

No. It is not in the NCERT Class 10 syllabus; it is enrichment material, useful for Olympiads.