Ceva’s Theorem

Ceva's Theorem

A theorem relating the way three concurrent cevians of a triangle divide the triangle's three sides.

Triangle ABC with cevians AD, BE, CF meeting at point inside; formula (AD/DB)·(BE/EC)·(CF/FA)=1

See also

Key Formula

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

Where:

$A, B, C$ = The vertices of the triangle
$D$ = The point where the cevian from vertex A meets side BC
$E$ = The point where the cevian from vertex B meets side CA
$F$ = The point where the cevian from vertex C meets side AB
$AF, FB, BD, DC, CE, EA$ = Directed or unsigned lengths of the segments into which the cevians divide the sides

Worked Example

Problem: In triangle ABC, cevians AD, BE, and CF are drawn. Point D lies on BC such that BD = 4 and DC = 6. Point E lies on CA such that CE = 3 and EA = 5. If the three cevians are concurrent, find the ratio AF : FB.

Step 1: Write down Ceva's Theorem for the three cevians.

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

Step 2: Substitute the known segment lengths.

\frac{AF}{FB} \cdot \frac{4}{6} \cdot \frac{3}{5} = 1

Step 3: Simplify the product of the two known ratios.

\frac{4}{6} \cdot \frac{3}{5} = \frac{12}{30} = \frac{2}{5}

Step 4: Solve for AF/FB by dividing both sides by 2/5.

\frac{AF}{FB} = \frac{1}{\,2/5\,} = \frac{5}{2}

Answer: AF : FB = 5 : 2. This means point F divides side AB so that AF is 2.5 times as long as FB.

Another Example

This example uses Ceva's Theorem in its converse direction — checking whether three given cevians meet at a single point, rather than finding an unknown ratio.

Problem: In triangle ABC, point D lies on BC with BD = 3 and DC = 3, point E lies on CA with CE = 2 and EA = 4, and point F lies on AB with AF = 8 and FB = 4. Determine whether cevians AD, BE, and CF are concurrent.

Step 1: Compute each ratio required by Ceva's Theorem.

\frac{AF}{FB} = \frac{8}{4} = 2, \quad \frac{BD}{DC} = \frac{3}{3} = 1, \quad \frac{CE}{EA} = \frac{2}{4} = \frac{1}{2}

Step 2: Multiply the three ratios together.

2 \cdot 1 \cdot \frac{1}{2} = 1

Step 3: Since the product equals 1, Ceva's Theorem confirms the cevians are concurrent.

Answer: Yes, the three cevians are concurrent because the product of the ratios equals 1.

Frequently Asked Questions

What is the difference between Ceva's Theorem and Menelaus's Theorem?

Both theorems involve ratios of segments on the sides of a triangle, but they address different situations. Ceva's Theorem deals with three cevians (lines from each vertex to the opposite side) being concurrent, and the product of the ratios equals +1. Menelaus's Theorem deals with three points on the sides (or their extensions) being collinear, and the product of the signed ratios equals −1.

When do you use Ceva's Theorem?

You use Ceva's Theorem whenever you need to prove that three cevians of a triangle meet at a single point, or when you know they are concurrent and want to find an unknown ratio in which one cevian divides its opposite side. It is especially useful in olympiad geometry and proofs involving special points like the centroid, incenter, or orthocenter.

Does Ceva's Theorem work if the cevian meets the extension of a side?

Yes, but you must use signed (directed) ratios. When a cevian meets the extension of a side rather than the side itself, the corresponding ratio is negative. The trigonometric form of Ceva's Theorem, which uses sines of angles instead of segment lengths, can also handle this situation cleanly.

Ceva's Theorem vs. Menelaus's Theorem

	Ceva's Theorem	Menelaus's Theorem
What it tests	Concurrency of three cevians (lines through vertices)	Collinearity of three points on the sides of a triangle
Product of ratios equals	+1 (using signed ratios)	−1 (using signed ratios)
Typical use case	Proving three lines meet at one point	Proving three points lie on one line
Lines involved	Each line passes through a vertex of the triangle	A single transversal line crosses all three sides (or extensions)

Why It Matters

Ceva's Theorem provides a single elegant condition for concurrency, replacing what might otherwise require coordinate geometry or multiple auxiliary constructions. It appears frequently in mathematical olympiads and competition geometry, and it also gives quick proofs that classical centers — the centroid, incenter, and symmedian point — exist. Understanding it strengthens your ability to work with ratios and proportional reasoning in triangle geometry.

Common Mistakes

Mistake: Writing the ratios in an inconsistent cyclic order around the triangle.

Correction: Always travel consistently around the triangle: each ratio should go in the same direction (e.g., AF/FB, then BD/DC, then CE/EA). Mixing directions gives an incorrect product.

Mistake: Ignoring signs when a cevian meets the extension of a side rather than the side itself.

Correction: If a cevian hits the extension of a side, the ratio for that side is negative. Use signed (directed) lengths so the product correctly equals +1 for concurrency.

Related Terms

Cevian — A line segment from a vertex to the opposite side
Concurrent — Three or more lines meeting at one point
Menelaus's Theorem — Analogous theorem for collinearity instead of concurrency
Triangle — The polygon to which Ceva's Theorem applies
Theorem — A proven mathematical statement
Side of a Polygon — The sides divided by the cevians
Cevian Theorem (Stewart's Theorem) — Relates a cevian's length to the side lengths