Difference between revisions of "Stewart's Theorem"
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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 + d^{2}m + d^{2}n </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 + d^{2}m + d^{2}n </math>. | ||
+ | However, since <math>m+n</math> can be written as a, we get the more common form: <math>cnc+bmb=man+dad</math> | ||
== See also == | == See also == |
Revision as of 08:51, 5 November 2006
Statement
![Stewart's theorem.png](https://wiki-images.artofproblemsolving.com//b/b3/Stewart%27s_theorem.png)
If a cevian of length d is drawn and divides side a into segments m and n, then
![$cnc + bmb = man + dad.$](http://latex.artofproblemsolving.com/0/2/c/02c0bb34d8c465ae29c1a5800bbc0524ea72bd1e.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 = d. 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: .
However, since
can be written as a, we get the more common form: