Difference between revisions of "Stewart's Theorem"

m (Reverted edits by Solafidefarms (Solafidefarms); changed back to last version by Agolsme)
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== Statement ==
 
== Statement ==
''(awaiting image)''<br>
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<center>[[Image:Stewart's_theorem.png]]</center>
If a [[cevian]] of length t is drawn and divides side a into segments m and n, then
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<br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br>
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If a [[cevian]] of length d is drawn and divides side a into segments m and n, then
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<center><math> cnc + bmb = man + dad.</math></center>
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== Proof ==
 
== Proof ==
For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{180 - \theta}</math>.
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For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{(180 - \theta)}</math>.
  
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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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 = d. We can write two equations:
*<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math>
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*<math> n^{2} + d^{2} - nd\cos{\angle CDA} = b^{2} </math>
*<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math>
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*<math> m^{2} + d^{2} + md\cos{\angle CDA} = c^{2} </math>
 
When we write everything in terms of cos(CDA) we have:
 
When we write everything in terms of cos(CDA) we have:
*<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math>
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*<math> \frac{n^2 + d^2 - b^2}{nd} = \cos{\angle CDA}</math>
*<math> \frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}</math>
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*<math> \frac{c^2 - m^2 -d^2}{md} = \cos{\angle CDA}</math>
 
 
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>
 
 
 
 
  
== Example ==
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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 + d^{2}m + d^{2}n </math>
''(awaiting addition)''
 
  
 
== See also ==  
 
== See also ==  
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
 
* [[Ceva's Theorem]]
 
* [[Ceva's Theorem]]
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* [[Geometry]]
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* [[Angle Bisector Theorem]]

Revision as of 21:35, 23 June 2006

Statement

Stewart's theorem.png

If a cevian of length d is drawn and divides side a into segments m and n, then

$cnc + bmb = man + dad.$


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 = d. We can write two equations:

  • $n^{2} + d^{2} - nd\cos{\angle CDA} = b^{2}$
  • $m^{2} + d^{2} + md\cos{\angle CDA} = c^{2}$

When we write everything in terms of cos(CDA) we have:

  • $\frac{n^2 + d^2 - b^2}{nd} = \cos{\angle CDA}$
  • $\frac{c^2 - m^2 -d^2}{md} = \cos{\angle CDA}$

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$

See also