Difference between revisions of "Stewart's Theorem"
Happyhuman (talk | contribs) (→Proof) |
Happyhuman (talk | contribs) (→Proof) |
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*<math>m^2n + n^2m = (m + n)mn = amn</math> and | *<math>m^2n + n^2m = (m + n)mn = amn</math> and | ||
*<math>d^2m + d^2n = d^2(m + n) = d^2a</math>. | *<math>d^2m + d^2n = d^2(m + n) = d^2a</math>. | ||
+ | <cmath>\text{center}</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. | ||
Revision as of 15:26, 7 January 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.