Difference between revisions of "Stewart's Theorem"
Happyhuman (talk | contribs) (→Proof) |
Happyhuman (talk | contribs) (→Proof) |
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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></math>m+n = a<math> so </math>m^2n + n^2m = (m + n)mn = amn<cmath> and </cmath>d^2m + d^2n = d^2(m + n) = d^2a<cmath>. | + | However, |
+ | <math></math>m+n = a<math> so </math>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 </cmath>c^2n + b^2m = amn + d^2a,<math></math> or Stewart's Theorem. | This simplifies our equation to yield </cmath>c^2n + b^2m = amn + d^2a,<math></math> or Stewart's Theorem. | ||
Revision as of 15:24, 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, $$ (Error compiling LaTeX. Unknown error_msg)m+n = am^2n + n^2m = (m + n)mn = amnd^2m + d^2n = d^2(m + n) = d^2ac^2n + b^2m = amn + d^2a,$$ (Error compiling LaTeX. Unknown error_msg) or Stewart's Theorem.