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>\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 CDA} = b^{2} </math>
+
*<math> n^{2} + d^{2} - 2\ce{nd}\cos{\angle CDA} = b^{2} </math>
*<math> m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2} </math>
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*<math> m^{2} + d^{2} - 2\ce{md}\cos{\angle ADB} = c^{2} </math>
  
 
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
 
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>
  
 
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>.
 
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>.
However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn</math> and we can rewrite this as <math>man + dad= bmb + cnc</math> (A man and his dad put a bomb in the sink). When you're practicing to memorize this formula, never practice it in the library or any other public place where other people can hear you.
+
However,  
 +
<math>m+n = a</math> so  
 +
<cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and  
 +
<cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath>
 +
This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</math> or Stewart's Theorem.
  
 
== See also ==  
 
== See also ==  

Revision as of 14:21, 28 October 2020

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices are $A$, $B$, $C$, 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 $man + dad = bmb + cnc$, a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")

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

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

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 \[m^2n + n^2m = (m + n)mn = amn\] and \[d^2m + d^2n = d^2(m + n) = d^2a.\] This simplifies our equation to yield $c^2n + b^2m = amn + d^2a,$ or Stewart's Theorem.

See also