# Difference between revisions of "User:Vqbc/Testing"

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− | + | == Statement == | |

+ | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] are <math>A</math>, <math>B</math>, <math>C</math>, respectively. If [[cevian]] <math>AD</math> is drawn so that <math>BD = m</math>, <math>DC = n</math> and <math>AD = d</math>, we have that <math>b^2m + c^2n = amn + d^2a</math>. (This is also often written <math>man + dad = bmb + cnc</math>, a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.") | ||

− | + | <center>[[Image:Stewart's_theorem.png]]</center> | |

− | + | == Proof == | |

+ | Applying the [[Law of Cosines]] in triangle <math>\triangle ABD</math> at [[angle]] <math>\angle ADB</math> and in triangle <math>\triangle ACD</math> at angle <math>\angle CDA</math>, we get the equations | ||

+ | *<math> n^{2} + d^{2} - 2\ce{nd}\cos{\angle CDA} = b^{2} </math> | ||

+ | *<math> m^{2} + d^{2} - 2\ce{md}\cos{\angle ADB} = c^{2} </math> | ||

− | === | + | Because angles <math>\angle ADB</math> and <math>\angle CDA</math> are [[supplementary]], <math>m\angle ADB = 180^\circ - m\angle CDA</math>. We can therefore solve both equations for the cosine term. Using the [[trigonometric identity]] <math>\cos{\theta} = -\cos{(180^\circ - \theta)}</math> gives us |

+ | *<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math> | ||

− | + | *<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math> | |

− | <math> | + | Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>. |

+ | However, | ||

+ | <math>m+n = a</math> so | ||

+ | <cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and | ||

+ | <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | ||

+ | This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</math> or Stewart's theorem. | ||

− | ==== | + | == See also == |

+ | * [[Menelaus' theorem]] | ||

+ | * [[Ceva's theorem]] | ||

+ | * [[Geometry]] | ||

+ | * [[Angle Bisector theorem]] | ||

− | + | [[Category:Geometry]] | |

− | + | [[Category:Theorems]] | |

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## Revision as of 23:27, 26 June 2021

## Statement

Given a triangle with sides of length opposite vertices are , , , respectively. If cevian is drawn so that , and , we have that . (This is also often written , a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")

## Proof

Applying the Law of Cosines in triangle at angle and in triangle at angle , we get the equations

Because angles and are supplementary, . We can therefore solve both equations for the cosine term. Using the trigonometric identity gives us

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: . However, so and This simplifies our equation to yield or Stewart's theorem.