Stewart's Theorem
Contents
Statement
(awaiting image)
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
(awaiting addition)