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| − | == Statement ==
| + | #REDIRECT[[Stewart's theorem]] |
| − | ''(awaiting image)''<br>
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| − | If a [[cevian]] of length t is drawn and divides side a into segments m and n, then
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| − | <br><center><math>c^{2}n + b^{2}m = (m+n)(t^{2} + mn)</math></center><br>
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| − | == Proof ==
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| − | For this proof we will use the law of cosines and the identity <math>\cos{\theta} = -\cos{180 - \theta}</math>.
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| − | 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:
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| − | *<math> n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2} </math>
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| − | *<math> m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2} </math>
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| − | When we write everything in terms of cos(CDA) we have:
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| − | *<math> \frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}</math>
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| − | *<math> \frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}</math>
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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 + t^{2}m + t^{2}n </math>
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| − | == Example ==
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| − | ''(awaiting addition)''
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| − | == See also ==
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| − | * [[Menelaus' Theorem]]
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| − | * [[Ceva's Theorem]]
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