# Difference between revisions of "Stewart's Theorem"

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== Statement == | == Statement == | ||

''(awaiting image)''<br> | ''(awaiting image)''<br> | ||

− | If a [[cevian]] of length | + | If a [[cevian]] of length t is drawn and divides side a into segments m and n, then |

− | <br><center><math> | + | <br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br> |

== Proof == | == Proof == | ||

− | ' | + | For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{180 - \theta}</math>. |

+ | |||

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

+ | *<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math> | ||

+ | *<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math> | ||

+ | When we write everything in terms of <math>\cos{\angle CDA}</math> we have: | ||

+ | *<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math> | ||

+ | *<math> \frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}</math> | ||

+ | |||

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

+ | |||

+ | |||

== Example == | == Example == |

## Revision as of 20:37, 18 June 2006

## Contents

## Statement

*(awaiting image)*

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

## Proof

For this proof we will use the law of cosines and the identity .

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

When we write everything in terms of we have:

Now we set the two equal and arrive at Stewart's theorem:

## Example

*(awaiting addition)*