Difference between revisions of "Stewart's Theorem"

 
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== Statement ==
 
== Statement ==
 
''(awaiting image)''<br>
 
''(awaiting image)''<br>
If a [[cevian]] of length d is drawn and divides side c into segments m and n, then
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If a [[cevian]] of length t is drawn and divides side a into segments m and n, then
<br><center><math>a^{2}n + b^{2}m = c(d^{2} + mn)</math></center><br>
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<br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br>
  
 
== Proof ==
 
== Proof ==
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For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{180 - \theta}</math>.
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Label the triangle <math>ABC</math> with a cevian extending from <math>A</math> onto <math>BC</math>, label that point <math>D</math>. Let CA = n Let DB = m. Let AD = t. We can write two equations:
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*<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math>
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*<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math>
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When we write everything in terms of <math>\cos{\angle CDA}</math> we have:
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*<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math>
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*<math> \frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}</math>
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Now we set the two equal and arrive at Stewart's theorem: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + t^{2}m + t^{2}n </math>
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== Example ==
 
== Example ==

Revision as of 20:37, 18 June 2006

Statement

(awaiting image)
If a cevian of length t is drawn and divides side a into segments m and n, then


$c^{2}n + b^{2}m = (m+n)(t^{2} + mn)$


Proof

For this proof we will use the law of cosines and the identity $\cos{\theta} = -\cos{180 - \theta}$.

Label the triangle $ABC$ with a cevian extending from $A$ onto $BC$, label that point $D$. Let CA = n Let DB = m. Let AD = t. We can write two equations:

  • $n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2}$
  • $m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2}$

When we write everything in terms of $\cos{\angle CDA}$ we have:

  • $\frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}$
  • $\frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}$

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + t^{2}m + t^{2}n$


Example

(awaiting addition)

See also

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