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Difference between revisions of "Stewart's Theorem"
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== Proof ==
 
== Proof ==
−
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
+
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>\angle CDA</math>, we get the equations
 
*<math> n^{2} + d^{2} - 2nd\cos{\angle ADB} = c^{2} </math>
 
*<math> n^{2} + d^{2} - 2nd\cos{\angle ADB} = c^{2} </math>
 
*<math> m^{2} + d^{2} - 2md\cos{\angle CDA} = b^{2} </math>
 
*<math> m^{2} + d^{2} - 2md\cos{\angle CDA} = b^{2} </math>
  
−
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
+
Because angles <math>\angle ADB</math> and <math>\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
 
*<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
 
*<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
 
*<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>
 
*<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>
Line 22: Line 22:
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Angle Bisector Theorem]]
 
* [[Angle Bisector Theorem]]
 +
 +
[[Category:Geometry]]
 +
 +
[[Category:Theorems]]

Revision as of 20:35, 14 October 2007

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 $\angle CDA$, 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 $\angle CDA$ 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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