# Difference between revisions of "Stewart's Theorem"

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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 | ||

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

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

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

## Latest revision as of 13:21, 28 October 2020

## 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.