Difference between revisions of "Stewart's Theorem"
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If a [[cevian]] of length t is drawn and divides side a into segments m and n, then | If a [[cevian]] of length t is drawn and divides side a into segments m and n, then | ||
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== Proof == | == Proof == |
Revision as of 18:54, 23 June 2006
Contents
Statement
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If a cevian of length t is drawn and divides side a into segments m and n, then
![$cmc+bmb=man+cnc$](http://latex.artofproblemsolving.com/5/b/d/5bdc24c81476684f8c11e53899ba3a17d1cc67c6.png)
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 cos(CDA) we have:
Now we set the two equal and arrive at Stewart's theorem:
Example
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