Difference between revisions of "Ceva's Theorem"
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== Statement == | == Statement == | ||
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A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that | A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that | ||
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br> | <br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br> | ||
where all segments in the formula are [[directed segments]]. | where all segments in the formula are [[directed segments]]. | ||
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+ | [[Image:Ceva1.PNG|center]] | ||
== Proof == | == Proof == | ||
− | + | Letting the [[altitude]] from <math>A</math> to <math>BC</math> have length <math>h</math> we have <math>[ABD]=\frac 12 BD\cdot h</math> and <math>[ACD]=\frac 12 DC\cdot h</math> where the brackets represent [[area]]. Thus <math>\frac{[ABD]}{[ACD]} = \frac{BD}{DC}</math>. In the same manner, we find that <math>\frac{[XBD]}{[XCD]} = \frac{BD}{DC}</math>. Thus <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}. </math></center> | |
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− | + | Likewise, we find that | |
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− | + | {| class="wikitable" style="margin: 1em auto 1em auto;height:100px" | |
− | + | | <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math> | |
− | <center><math>\frac{ | + | |- |
− | + | | <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math> | |
− | + | |} | |
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− | + | Thus <center><math> \frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1 \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB. </math></center> | |
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− | + | <math>\mathcal{QED}</math> | |
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− | <center><math>\frac{ | ||
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− | == | + | == Examples == |
− | Suppose AB, AC, and BC have lengths 13, 14, and 15. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>. Find BD and DC.<br> | + | # Suppose AB, AC, and BC have lengths 13, 14, and 15. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>. Find BD and DC.<br> <br> If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>. From this, we find <math>x = 12</math> and <math>y = 3</math>. |
− | <br> | + | # See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]]. |
− | If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>. From this, we find <math>x = 12</math> and <math>y = 3</math>. | + | # See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]]. |
== See also == | == See also == | ||
* [[Menelaus' Theorem]] | * [[Menelaus' Theorem]] | ||
* [[Stewart's Theorem]] | * [[Stewart's Theorem]] |
Revision as of 10:03, 18 August 2006
Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.
Contents
Statement
A necessary and sufficient condition for where and are points of the respective side lines of a triangle , to be concurrent is that
where all segments in the formula are directed segments.
Proof
Letting the altitude from to have length we have and where the brackets represent area. Thus . In the same manner, we find that . Thus
Likewise, we find that
Thus
Examples
- Suppose AB, AC, and BC have lengths 13, 14, and 15. If and . Find BD and DC.
If and , then , and . From this, we find and . - See the proof of the concurrency of the altitudes of a triangle at the orthocenter.
- See the proof of the concurrency of the perpendicual bisectors of a triangle at the circumcenter.