Difference between revisions of "Stewart's theorem"
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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 | ||
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<cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | <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. | This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</math> or Stewart's theorem. | ||
+ | |||
+ | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ||
+ | https://youtu.be/jEVMgWKQIW8 | ||
+ | |||
+ | ~IceMatrix | ||
== See also == | == See also == |
Revision as of 16:15, 11 September 2021
Contents
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.
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix