Difference between revisions of "Ceva's Theorem"
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<math>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</math> | <math>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</math> | ||
− | + | Dividing by <math>xyz</math> yields: | |
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<math>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</math>, which is equivalent to Ceva's theorem | <math>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</math>, which is equivalent to Ceva's theorem |
Revision as of 21:59, 1 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 concur 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.