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>. | ||
− | However, <math>m+n = a</math> so < | + | 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 == | == See also == |
Revision as of 14:21, 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.