# Difference between revisions of "Ceva's Theorem"

(→Statement: reciprocal, image) |
m (→Proof) |
||

(25 intermediate revisions by 14 users not shown) | |||

Line 5: | Line 5: | ||

[[Image:Ceva1.PNG|thumb|right]] | [[Image:Ceva1.PNG|thumb|right]] | ||

− | 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]] 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>, | ||

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

− | where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] | + | where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] of each of the ratios, as the reciprocal of <math>1</math> is <math>1</math>. |

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

Line 19: | Line 22: | ||

First, suppose <math>AD, BE, CF </math> meet at a point <math>X </math>. We note that triangles <math>ABD, ADC </math> have the same altitude to line <math>BC </math>, but bases <math>BD </math> and <math>DC </math>. It follows that <math> \frac {BD}{DC} = \frac{[ABD]}{[ADC]} </math>. The same is true for triangles <math>XBD, XDC </math>, so | First, suppose <math>AD, BE, CF </math> meet at a point <math>X </math>. We note that triangles <math>ABD, ADC </math> have the same altitude to line <math>BC </math>, but bases <math>BD </math> and <math>DC </math>. It follows that <math> \frac {BD}{DC} = \frac{[ABD]}{[ADC]} </math>. The same is true for triangles <math>XBD, XDC </math>, so | ||

+ | |||

<center><math> \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} </math>. </center> | <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} </math>. </center> | ||

Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>, | Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>, | ||

Line 26: | Line 30: | ||

</center> | </center> | ||

− | 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]]== | ||

+ | |||

+ | Since <math>D\in BC</math>, we can write its coordinates as <math>(0,d,1-d)</math>. The equation of line <math>AD</math> is then <math>z=\frac{1-d}{d}y</math>. | ||

+ | |||

+ | Similarly, since <math>E=(1-e,0,e)</math>, and <math>F=(f,1-f,0)</math>, we can see that the equations of <math>BE</math> and <math>CF</math> respectively are <math>x=\frac{1-e}{e}z</math> and <math>y=\frac{1-f}{f}x</math> | ||

+ | |||

+ | [[Multiplying]] the three together yields the solution to the equation: | ||

+ | |||

+ | <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: | ||

+ | |||

+ | |||

+ | <math>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</math>, which is equivalent to Ceva's theorem | ||

+ | |||

+ | QED | ||

== Trigonometric Form == | == Trigonometric Form == | ||

Line 52: | Line 73: | ||

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

− | == | + | == Problems == |

+ | ===Introductory=== | ||

+ | *Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>, find <math>BD</math> and <math>DC</math>. ([[Ceva's Theorem/Problems|Source]]) | ||

− | + | ===Intermediate=== | |

− | + | *In <math>\Delta ABC, AD, BE, CF</math> are concurrent lines. <math>P, Q, R</math> are points on <math>EF, FD, DE</math> such that <math>DP, EQ, FR</math> are concurrent. Prove that (using ''plane geometry'') <math>AP, BQ, CR</math> are concurrent. (<url>viewtopic.php?f=151&t=543574 </url>) | |

− | |||

== See also == | == See also == | ||

+ | * [[Stewart's Theorem]] | ||

* [[Menelaus' Theorem]] | * [[Menelaus' Theorem]] | ||

− | |||

[[Category:Geometry]] | [[Category:Geometry]] | ||

[[Category:Theorems]] | [[Category:Theorems]] | ||

− | |||

− |

## Latest revision as of 02:35, 24 July 2020

**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 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>)