Difference between revisions of "Stewart's Theorem"
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== Statement == | == Statement == | ||
''(awaiting image)''<br> | ''(awaiting image)''<br> | ||
| − | If a [[cevian]] of length | + | 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 == | ||
| − | ' | + | For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{180 - \theta}</math>. |
| + | |||
| + | Label the triangle <math>ABC</math> with a cevian extending from <math>A</math> onto <math>BC</math>, label that point <math>D</math>. Let CA = n Let DB = m. Let AD = t. We can write two equations: | ||
| + | *<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math> | ||
| + | *<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math> | ||
| + | When we write everything in terms of <math>\cos{\angle CDA}</math> we have: | ||
| + | *<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> | ||
| + | |||
| + | 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 + t^{2}m + t^{2}n </math> | ||
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== Example == | == Example == | ||
Revision as of 20:37, 18 June 2006
Contents
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 
we have:
Now we set the two equal and arrive at Stewart's theorem: 
Example
(awaiting addition)
