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| − | '''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].
| + | #REDIRECT[[Ceva's theorem]] |
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| − | == Statement ==
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| − | [[Image:Ceva1.PNG|thumb|right]]
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| − | 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> [[concur]] iff (if and only if)
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| − | <br><center>
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| − | <math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
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| − | </center><br>
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| − | where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] or each of the ratios, as the reciprocal of <math>1</math> is <math>1</math>.
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| − | (Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
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| − | == Proof ==
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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>.
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| − | 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
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| − | <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} </math>. </center>
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| − | Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>,
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| − | so
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| − | <center>
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| − | <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>.
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| − | </center>
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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> concurrs with <math>AD </math> and <math>BE </math>. {{Halmos}}
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| − | ==Partial proof by Barycentric Coordinates==
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| − | Let three said cevians intersect at point <math>O</math>. Let the barycentric coordinates of <math>O</math> be <math>(X,Y,Z)</math>. Then, because the coordinates are [[homogeneous]], we can say <math>AE=AF=X</math>, <math>BF=BD=Y</math>, and <math>CD=CE=Z</math>. Then, plugging the values into the equation yields:
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| − | <math>\frac{Y}{Z} \cdot \frac{Z}{X}\cdot \frac{X}{Y} = 1 \shadedbox</math>
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| − | == Trigonometric Form ==
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| − | The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians <math>AD,BE,CF</math> concur if and only if
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| − | <center>
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| − | <math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.</math>
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| − | </center>
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| − | === Proof ===
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| − | First, suppose <math>AD, BE, CF </math> concur at a point <math>X </math>. We note that
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| − | <center>
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| − | <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>
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| − | and similarly,
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| − | <center>
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| − | <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>
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| − | It follows that
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| − | <center>
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| − | <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>.
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| − | </center>
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| − | Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math>\pi </math> to be either positive or negative.
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| − | The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. {{Halmos}}
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| − | == Problems ==
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| − | ===Introductory===
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| − | *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]])
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| − | ===Intermediate===
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| − | ===Olympiad===
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| − | ==Other Notes==
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| − | *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.
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| − | * The existence of [[isotomic conjugate]]s can be shown by classic Ceva, and the existence of [[isogonal conjugate]]s can be shown by trig Ceva.
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| − | == See also ==
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| − | * [[Menelaus' Theorem]]
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| − | [[Category:Geometry]]
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| − | [[Category:Theorems]]
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