Difference between revisions of "Stewart's theorem"
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<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>man + dad = bmb + cnc,</math> or Stewart's theorem. | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. | ||
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== Proof 2 (Pythagorean Theorem) == | == Proof 2 (Pythagorean Theorem) == |
Latest revision as of 12:03, 25 July 2024
Contents
Statement
Given a triangle with sides of length
and opposite vertices
,
,
, respectively. If cevian
is drawn so that
,
and
, we have that
. (This is also often written
, a phrase which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.") That is Stewart's Theorem. I know, it's easy to memorize.
![Stewart's theorem.png](https://wiki-images.artofproblemsolving.com//b/b3/Stewart%27s_theorem.png)
Proof 1
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.
Proof 2 (Pythagorean Theorem)
Let the altitude from to
meet
at
. Let
,
, and
. So, applying Pythagorean Theorem on
yields
Since ,
Applying Pythagorean on yields
Substituting ,
, and
in
and
gives
Notice that
are equal to each other. Thus,
Rearranging the equation gives Stewart's Theorem:
~sml1809
Proof 3 (Barycentrics)
Let the following points have the following coordinates:
Our displacement vector has coordinates
. Plugging this into the barycentric distance formula, we obtain
Multiplying by
, we get
. Substituting
with
, we find Stewart's Theorem:
~kn07
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix