# Difference between revisions of "Ceva's Theorem"

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− | Let <math>ABC </math> be a triangle, and let <math>D, E, F </math> be points on lines <math>BC, CA, AB </math>, respectively. Lines <math>AD, BE, CF </math> [[ | + | Let <math>ABC </math> be a triangle, and let <math>D, E, F </math> be points on lines <math>BC, CA, AB </math>, respectively. Lines <math>AD, BE, CF </math> are [[concurrent]] iff (if and only if) |

<br><center> | <br><center> | ||

<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>, | <math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>, |

## Revision as of 09:33, 11 June 2013

**Ceva's Theorem** is a criterion for the concurrence of cevians in a triangle.

## Contents

## Statement

Let be a triangle, and let be points on lines , respectively. Lines are concurrent iff (if and only if)

,

where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of is .

(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)

The proof using Routh's Theorem is extremely trivial, so we will not include it.

## Proof

We will use the notation to denote the area of a triangle with vertices .

First, suppose meet at a point . We note that triangles have the same altitude to line , but bases and . It follows that . The same is true for triangles , so

Similarly, and , so

.

Now, suppose satisfy Ceva's criterion, and suppose intersect at . Suppose the line intersects line at . We have proven that must satisfy Ceva's criterion. This means that

so

and line concurrs with and . ∎

## Proof by Barycentric coordinates

Since , we can write its coordinates as . The equation of line is then .

Similarly, since , and , we can see that the equations of and respectively are and

Multiplying the three together yields the solution to the equation:

Dividing by yields:

, which is equivalent to Ceva's theorem

QED

## Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians concur if and only if

### Proof

First, suppose concur at a point . We note that

and similarly,

It follows that

.

Here, sign is irrelevant, as we may interpret the sines of directed angles mod to be either positive or negative.

The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. ∎

## Problems

### Introductory

- Suppose , and have lengths , and , respectively. If and , find and . (Source)

### Intermediate

### Olympiad

## Other Notes

- The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicular bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existence of the centroid can be shown by Ceva, and the existence of the incenter can be shown using trig Ceva. However, there are more elegant methods for proving each of these results, and in any case, any result obtained by Ceva's Theorem can be obtained using ratios of areas.
- The existence of isotomic conjugates can be shown by classic Ceva, and the existence of isogonal conjugates can be shown by trig Ceva.