Difference between revisions of "Stewart's Theorem"
(weird angle latex problem) |
|||
Line 10: | Line 10: | ||
*<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
Contents
[hide]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)