Difference between revisions of "Ceva's Theorem"
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− | Now, suppose <math>D, E,F </math> satisfy Ceva's criterion, and suppose <math>AD, BE </math> intersect at <math>X </math>. Suppose the line <math>CX </math> intersects line <math>AB </math> at <math>F' </math>. We have proven that <math>F' </math> must satisfy Ceva's criterion. This means that <center><math> \frac{AF'}{F'B} = \frac{AF}{FB} </math>, </center> so <center><math>F' = F </math>, </center> and line <math>CF </math> | + | Now, suppose <math>D, E,F </math> satisfy Ceva's criterion, and suppose <math>AD, BE </math> intersect at <math>X </math>. Suppose the line <math>CX </math> intersects line <math>AB </math> at <math>F' </math>. We have proven that <math>F' </math> must satisfy Ceva's criterion. This means that <center><math> \frac{AF'}{F'B} = \frac{AF}{FB} </math>, </center> so <center><math>F' = F </math>, </center> and line <math>CF </math> concurs with <math>AD </math> and <math>BE </math>. {{Halmos}} |
==Proof by [[Barycentric coordinates]]== | ==Proof by [[Barycentric coordinates]]== |
Revision as of 01:35, 24 July 2020
Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.
Contents
[hide]Statement
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.
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 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
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
- In are concurrent lines. are points on such that are concurrent. Prove that (using plane geometry) are concurrent. (<url>viewtopic.php?f=151&t=543574 </url>)