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>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn</math> and we can rewrite this as <math>man + dad= bmb + cnc</math> (A man and his dad put a bomb in the sink). | + | However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn</math> and we can rewrite this as <math>man + dad= bmb + cnc</math> (A man and his dad put a bomb in the sink). When you're practicing to memorize this formula, never practice it in the library or any other public place where other people can hear you. |
== See also == | == See also == | ||
Revision as of 17:10, 31 March 2018
Statement
Given a triangle 
with sides of length 
opposite vertices 
, 
, 
, respectively. If cevian 
is drawn so that 
, 
and 
, we have that 
. (This is also often written 
, a form which invites mnemonic memorization, e.g. "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 we can rewrite this as 
(A man and his dad put a bomb in the sink). When you're practicing to memorize this formula, never practice it in the library or any other public place where other people can hear you.
