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| − | == Statement ==
| + | #REDIRECT[[Stewart's theorem]] |
| − | <center>[[Image:Stewart's_theorem.png]]</center>
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| − | If a [[cevian]] of length d is drawn and divides side a into segments m and n, then
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| − | <center><math> cnc + bmb = man + dad.</math></center>
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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 = d. We can write two equations:
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| − | *<math> n^{2} + d^{2} - nd\cos{\angle CDA} = b^{2} </math>
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| − | *<math> m^{2} + d^{2} + md\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 + d^2 - b^2}{nd} = \cos{\angle CDA}</math>
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| − | *<math> \frac{c^2 - m^2 -d^2}{md} = \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 + d^{2}m + d^{2}n </math>
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| − | == See also ==
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| − | * [[Menelaus' Theorem]]
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| − | * [[Ceva's Theorem]]
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| − | * [[Geometry]]
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| − | * [[Angle Bisector Theorem]]
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