Ceva's theorem is a criterion for the concurrence of cevians in a triangle.
Let be a triangle, and let be points on lines , respectively. Lines are concurrent if and only if
where lengths are directed. This also works for the reciprocal of 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.
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
and line concurs 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
The trigonometric form of Ceva's theorem states that cevians concur if and only if
First, suppose concur at a point . We note that
It follows that
Here, the 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. ∎
- Suppose , and have lengths , and , respectively. If and , find and . (Source)
- In are concurrent lines. are points on such that are concurrent. Prove that (using plane geometry) are concurrent.
- Let be the midpoint of side of triangle . Points and lie on line segments and , respectively, such that and are parallel. Point lies on line segment . Lines and intersect at and lines and meet at . Prove that are collinear. (Ceva I.2)