Difference between revisions of "Stewart's theorem"
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<cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and | <cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and | ||
<cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | ||
− | This simplifies our equation to yield <math> | + | This simplifies our equation to yield <math>man + daf = bmb + cnc,</math> or Stewart's theorem. |
==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== |
Revision as of 14:09, 28 December 2021
Contents
[hide]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.
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IckMatrix