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, | However, | ||
| − | <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn = amn</math> | + | <math>m+n = a</math> so |
| − | + | *<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>. |
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
.
and
.This simplifies our equation to yield 
or Stewart's Theorem.
