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Difference between revisions of "Stewart's Theorem"

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== Statement ==
 
== Statement ==
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Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] <math>A</math>, <math>B</math>, <math>C</math>, respectively.  If [[cevian]] <math>AD</math> is drawn so that <math>BD = m</math>, <math>DC = n</math> and <math>AD = d</math>, we have that <math>b^2m + c^2n = amn + d^2a</math>.  (This is also often written <math> cnc + bmb = man + dad</math>, a form which invites mnemonic memorization.)
 +
 
<center>[[Image:Stewart's_theorem.png]]</center>
 
<center>[[Image:Stewart's_theorem.png]]</center>
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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>
 
  
  
 
== 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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Applying the [[Law of Cosines]] in triangle <math>\triangle ABD</math> at [[angle]] <math>\angle ADB</math> and in triangle <math>\triangle ACD</math> at angle <math>\displaystyle \angle CDA</math>, we get the equations
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*<math> n^{2} + d^{2} - 2nd\cos{\angle ADB} = c^{2} </math>
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*<math> m^{2} + d^{2} - 2md\cos{\angle CDA} = b^{2} </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 = d. We can write two equations:
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Because angles <math>\angle ADB</math> and <math>\displaystyle \angle CDA</math> are [[supplementary]], <math>m\angle ADB = 180^\circ - m\angle CDA</math>. We can therefore solve both equations for the cosine term.  Using the [[trigonometric identity]] <math>\cos{\theta} = -\cos{(180^\circ - \theta)}</math> gives us
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*<math> n^{2} + d^{2} - nd\cos{\angle CDA} = b^{2} </math>
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*<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
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*<math> m^{2} + d^{2} + md\cos{\angle CDA} = c^{2} </math>
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*<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>
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When we write everything in terms of cos(CDA) we have:
 
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*<math> \frac{n^2 + d^2 - b^2}{nd} = \cos{\angle CDA}</math>
 
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*<math> \frac{c^2 - m^2 -d^2}{md} = \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 + d^{2}m + d^{2}n </math>.
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Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>.
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However, since <math>m+n</math> can be written as a, we get the more common form: <math>cnc+bmb=man+dad</math>
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However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn</math> and we can rewrite this as <math>c^2n + b^2m = amn + d^2a</math>.
  
 
== See also ==  
 
== See also ==  

Revision as of 14:11, 5 November 2006

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices $A$, $B$, $C$, respectively. If cevian $AD$ is drawn so that $BD = m$, $DC = n$ and $AD = d$, we have that $b^2m + c^2n = amn + d^2a$. (This is also often written $cnc + bmb = man + dad$, a form which invites mnemonic memorization.)

Stewart's theorem.png


Proof

Applying the Law of Cosines in triangle $\triangle ABD$ at angle $\angle ADB$ and in triangle $\triangle ACD$ at angle $\displaystyle \angle CDA$, we get the equations

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

Because angles $\angle ADB$ and $\displaystyle \angle CDA$ are supplementary, $m\angle ADB = 180^\circ - m\angle CDA$. We can therefore solve both equations for the cosine term. Using the trigonometric identity $\cos{\theta} = -\cos{(180^\circ - \theta)}$ gives us

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

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$. However, $m+n = a$ so $m^2n + n^2m = (m + n)mn$ and we can rewrite this as $c^2n + b^2m = amn + d^2a$.

See also

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