# Difference between revisions of "Stewart's Theorem"

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− | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] <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, e | + | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] <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> | <center>[[Image:Stewart's_theorem.png]]</center> |

## Revision as of 15:51, 17 June 2018

## Statement

Given a triangle with sides of length opposite vertices , , , 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 we can rewrite this as (A man and his dad put a bomb in the sink). When you're practicing to memorize this formula, never practice it in the library or any other public place where other people can hear you.