Difference between revisions of "Stewart's Theorem"
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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,
$$ (Error compiling LaTeX. Unknown error_msg)m+n = a
m^2n + n^2m = (m + n)mn = amn
d^2m + d^2n = d^2(m + n) = d^2a
c^2n + b^2m = amn + d^2a,$$ (Error compiling LaTeX. Unknown error_msg) or Stewart's Theorem.