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> | + | 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> | + | 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> | ||
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* [[Geometry]] | * [[Geometry]] | ||
* [[Angle Bisector Theorem]] | * [[Angle Bisector Theorem]] | ||
+ | |||
+ | [[Category:Geometry]] | ||
+ | |||
+ | [[Category:Theorems]] |
Revision as of 21:35, 14 October 2007
Statement
Given a triangle with sides of length
opposite vertices
,
,
, respectively. If cevian
is drawn so that
,
and
, we have that
. (This is also often written
, a form which invites mnemonic memorization.)
![Stewart's theorem.png](https://wiki-images.artofproblemsolving.com//b/b3/Stewart%27s_theorem.png)
Proof
Applying the Law of Cosines in triangle at angle
and in triangle
at angle
, we get the equations
Because angles and
are supplementary,
. We can therefore solve both equations for the cosine term. Using the trigonometric identity
gives us
Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: .
However,
so
and we can rewrite this as
.