Stewart's Theorem: Difference between revisions
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*<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math> | *<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math> | ||
*<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math> | *<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math> | ||
When we write everything in terms of | When we write everything in terms of cos(CDA) we have: | ||
*<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math> | *<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> | *<math> \frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}</math> | ||
Revision as of 20:38, 18 June 2006
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 cos(CDA) we have:
Now we set the two equal and arrive at Stewart's theorem: 
Example
(awaiting addition)
