Difference between revisions of "Stewart's theorem"
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<math>C: (0,0,1)</math> | <math>C: (0,0,1)</math> | ||
− | <math>D: (0, \frac{n}{m+n},\frac{m}{m+n})</math> | + | <math>D: \left(0, \frac{n}{m+n},\frac{m}{m+n}\right)</math> |
− | Our displacement vector <math>\overrightarrow{AD}</math> has coordinates <math>(1, -\frac{n}{m+n}, -\frac{m}{m+n})</math>. Plugging this into the barycentric distance formula, we obtain <cmath>d^2=-(m+n)^2 \left(\frac{mn}{(m+n)^2} \right)-b^2 \left ( -\frac{m}{m+n} \right)-c^2 \left(-\frac{n}{m+n}\right)=-mn+\frac{b^2m+c^2n}{m+n}</cmath> Multiplying by <math>m+n</math>, we get <math>d^2(m+n)+mn(m+n)=b^2m+c^2n</math>. Substituting <math>m+n</math> with <math>a</math>, we find Stewart's Theorem: <cmath>\boxed{d^2a+amn=b^2m+c^2n}</cmath> | + | Our displacement vector <math>\overrightarrow{AD}</math> has coordinates <math>\left(1, -\frac{n}{m+n}, -\frac{m}{m+n}\right)</math>. Plugging this into the barycentric distance formula, we obtain <cmath>d^2=-(m+n)^2 \left(\frac{mn}{(m+n)^2} \right)-b^2 \left ( -\frac{m}{m+n} \right)-c^2 \left(-\frac{n}{m+n}\right)=-mn+\frac{b^2m+c^2n}{m+n}</cmath> Multiplying by <math>m+n</math>, we get <math>d^2(m+n)+mn(m+n)=b^2m+c^2n</math>. Substituting <math>m+n</math> with <math>a</math>, we find Stewart's Theorem: <cmath>\boxed{d^2a+amn=b^2m+c^2n}</cmath> |
~kn07 | ~kn07 |
Revision as of 20:33, 27 January 2023
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.
Good Job! You mastered 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