Difference between revisions of "Ceva's Theorem"
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We will use the notation <math>[ABC]</math> to denote the area of a triangle with vertices <math>A,B,C</math>. | 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 | + | 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 |
<cmath>\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}</cmath> | <cmath>\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}</cmath> | ||
Similarly, <math>\frac{CE}{EA} = \frac{[BCX]}{[BXA]}</math> and <math>\frac{AF}{FB} = \frac{[CAX]}{[CXB]}</math>, so | Similarly, <math>\frac{CE}{EA} = \frac{[BCX]}{[BXA]}</math> and <math>\frac{AF}{FB} = \frac{[CAX]}{[CXB]}</math>, so | ||
<cmath>\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1</cmath> | <cmath>\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1</cmath> | ||
− | 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 < | + | 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 <cmath>\frac{AF'}{F'B} = \frac{AF}{FB}</cmath> so <cmath>F' = F</cmath> and line <math>CF</math> concurs with <math>AD</math> and <math>BE</math>. <math>\square</math> |
== Proof by [[Barycentric coordinates]] == | == Proof by [[Barycentric coordinates]] == | ||
− | Since < | + | 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 < | + | 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: | Multiplying the three together yields the solution to the equation: | ||
<cmath>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</cmath> | <cmath>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</cmath> | ||
− | Dividing by < | + | Dividing by <math>xyz</math> yields: |
<cmath>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</cmath>, which is equivalent to Ceva's Theorem. | <cmath>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</cmath>, which is equivalent to Ceva's Theorem. | ||
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== Trigonometric Form == | == Trigonometric Form == | ||
− | The [[trigonometry|trigonometric]] form of Ceva's Theorem states that cevians < | + | The [[trigonometry|trigonometric]] form of Ceva's Theorem states that cevians <math>AD,BE,CF</math> concur if and only if |
<cmath>\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1</cmath> | <cmath>\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1</cmath> | ||
=== Proof === | === Proof === | ||
− | First, suppose < | + | First, suppose <math>AD, BE, CF </math> concur at a point <math>X </math>. We note that |
<cmath>\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}</cmath> | <cmath>\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}</cmath> | ||
and similarly, | and similarly, | ||
<cmath>\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}</cmath> | <cmath>\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}</cmath> | ||
It follows that | It follows that | ||
− | <cmath> \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} </cmath> <br> <cmath> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1</cmath> | + | <cmath>\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} </cmath> <br> <cmath>\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1</cmath> |
− | Here, the sign is irrelevant, as we may interpret the sines of [[directed angles]] mod < | + | Here, the 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. < | + | The converse follows by an argument almost identical to that used for the first form of Ceva's theorem. <math>\square</math> |
== Problems == | == Problems == | ||
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=== Introductory === | === Introductory === | ||
− | * Suppose < | + | * 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 === | === Intermediate === | ||
− | *In < | + | *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. |
− | *Let < | + | *Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> lie on line segments <math>BC</math> and <math>CA</math>, respectively, such that <math>DE</math> and <math>AB</math> are parallel. Point <math>P</math> lies on line segment <math>AM</math>. Lines <math>EM</math> and <math>CP</math> intersect at <math>X</math> and lines <math>DP</math> and <math>CM</math> meet at <math>Y</math>. Prove that <math>X,Y,B</math> are collinear. ([[Ceva's Theorem/Problems|Source]]) |
== See Also == | == See Also == |
Latest revision as of 21:11, 28 April 2025
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.
Trigonometric Form
The trigonometric form of Ceva's Theorem states that cevians concur if and only if
Proof
First, suppose concur at a point
. We note that
and similarly,
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.
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.
- 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. (Source)