Difference between revisions of "Stewart's Theorem"

Revision as of 17:54, 23 June 2006

Statement

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If a cevian of length t is drawn and divides side a into segments m and n, then

$cmc+bmb=man+cnc$

Proof

For this proof we will use the law of cosines and the identity $\cos{\theta} = -\cos{180 - \theta}$ .

Label the triangle $ABC$ with a cevian extending from $A$ onto $BC$ , label that point $D$ . Let CA = n Let DB = m. Let AD = t. We can write two equations:

$n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2}$
$m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2}$

When we write everything in terms of cos(CDA) we have:

$\frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}$
$\frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}$

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + t^{2}m + t^{2}n$

Example

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@@ Line 2: / Line 2: @@
 ''(awaiting image)''<br>
 If a [[cevian]] of length t is drawn and divides side a into segments m and n, then
-<br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br>
+<br><center><math>cmc+bmb=man+cnc</math></center><br>
 == Proof ==

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Difference between revisions of "Stewart's Theorem"

Revision as of 17:54, 23 June 2006

Contents

Statement

Proof

Example

See also