# Difference between revisions of "Stewart's theorem"

Etmetalakret (talk | contribs) (Created page with "== Statement == Given a triangle <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite vertices are <math>A</math>, <math>B</math>, <m...") |
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== Proof == | == 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 | 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} - | + | *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math> |

− | *<math> m^{2} + d^{2} - | + | *<math> m^{2} + d^{2} - 2md\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 | 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 |

## Revision as of 14:49, 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.