Difference between revisions of "Ceva's Theorem"
m (proofreading) |
m (→Proof by Barycentric coordinates) |
||
(37 intermediate revisions by 17 users not shown) | |||
Line 1: | Line 1: | ||
− | '''Ceva's Theorem''' is | + | '''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]]. |
== Statement == | == Statement == | ||
− | + | ||
− | + | [[Image:Ceva1.PNG|thumb|right]] | |
− | <br><center><math>BD | + | 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 |
− | where | + | <br><center> |
+ | <math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>, | ||
+ | </center><br> | ||
+ | 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.) | ||
+ | |||
+ | |||
+ | The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it. | ||
== Proof == | == Proof == | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | < | + | We will use the notation <math>[ABC] </math> to denote the area of a triangle with vertices <math>A,B,C </math>. |
+ | |||
+ | 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> | ||
+ | Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>, | ||
+ | so | ||
+ | <center> | ||
+ | <math> \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1 </math>. | ||
+ | </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> concurrs 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 == | ||
+ | |||
+ | The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians <math>AD,BE,CF</math> concur if and only if | ||
+ | <center> | ||
+ | <math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.</math> | ||
+ | </center> | ||
+ | |||
+ | === Proof === | ||
+ | |||
+ | First, suppose <math>AD, BE, CF </math> concur at a point <math>X </math>. We note that | ||
+ | <center> | ||
+ | <math> \frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} </math>, </center> | ||
+ | and similarly, | ||
+ | <center> | ||
+ | <math> \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center> | ||
+ | It follows that | ||
+ | <center> | ||
+ | <math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math> <br> <br> <math> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 </math>. | ||
+ | </center> | ||
+ | |||
+ | Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math>\pi </math> 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. {{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:Theorems]] |
Revision as of 18:15, 6 August 2018
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 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
- 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>)