Difference between revisions of "Stewart's Theorem"

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== Statement ==
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#REDIRECT[[Stewart's theorem]]
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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>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br>
 
 
 
== Proof ==
 
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:
 
*<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math>
 
*<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math>
 
When we write everything in terms of <math>\cos{\angle CDA}</math> we have:
 
*<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math>
 
*<math> \frac{c^2 - m^2 -t^2}{mt} = \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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== See also ==
 
* [[Menelaus' Theorem]]
 
* [[Ceva's Theorem]]
 

Latest revision as of 15:22, 9 May 2021

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