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| − | '''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].
| + | #REDIRECT[[Ceva's theorem]] |
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| − | == 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
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| − | <br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
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| − | where all segments in the formula are [[directed segments]].
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| − | [[Image:Ceva1.PNG|center]]
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| − | == Proof ==
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| − | 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"
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| − | | <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math>
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| − | | <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math>
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| − | |}
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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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| − | == Alternate Formulation ==
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| − | The [[trig]] version of Ceva's Theorem states that cevians <math>AD,BE,CF</math> are concurrent if and only if
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| − | <center>$\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.$</center>
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| − | === Proof ===
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| − | ''This proof is incomplete. If you can finish it, please do so. Thanks!''
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| − | We will use Ceva's Theorem in the form that was already proven to be true.
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| − | First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem. The [[Law of Sines]] tells us that <center>$\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.$</center>
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| − | Likewise, we get
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| − | {| class="wikitable" style="margin: 1em auto 1em auto"
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| − | | $\sin ACF = \frac{AF}{AC\sin CFA}$
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| − | | $\sin CBE = \frac{CE}{BC\sin BEC}$
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| − | | $\sin CAD = \frac{CD}{AC\sin ADC}$
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| − | | $\sin BCF = \frac{BF}{BC\sin BFC}$
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| − | | $\sin ABE = \frac{AE}{AB\sin AEB}$
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| − | |}
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| − | Thus
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| − | {| class="wikitable"
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| − | | <math>\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE</math>
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| − | | $\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB}
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| − | |}
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| − | == Examples ==
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| − | # 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>.
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| − | # See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]].
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| − | # See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]].
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| − | == See also ==
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| − | * [[Menelaus' Theorem]]
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| − | * [[Stewart's Theorem]]
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