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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]] are <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>man + dad = bmb + cnc</math>, a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
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Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] are <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>man + dad = bmb + cnc</math>, a form which invites mnemonic memorization, i.e. "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.)
  
 
<center>[[Image:Stewart's_theorem.png]]</center>
 
<center>[[Image:Stewart's_theorem.png]]</center>

Revision as of 19:16, 14 February 2019

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." 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.)

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

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

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 $m^2n + n^2m = (m + n)mn$ and we can rewrite this as $man + dad= bmb + cnc$ (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.

See also

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