Difference between revisions of "Stewart's Theorem"
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m (Reverted edits by Solafidefarms (Solafidefarms); changed back to last version by Agolsme) |
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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 | ||
− | <br><center><math> | + | <br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br> |
== Proof == | == Proof == |
Revision as of 18:55, 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
![$c^{2}n + b^{2}m = (m+n)(t^{2} + mn)$](http://latex.artofproblemsolving.com/5/8/4/584ba47a165beef167e4482a2f12ad9b242c296b.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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