Difference between revisions of "Stewart's theorem"
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== Statement == | == Statement == | ||
− | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] | + | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> and opposite [[vertex | vertices]] <math>A</math>, <math>B</math>, <math>C</math>, respectively. If [[cevian]] <math>AD</math> is drawn so that <math>BD = m</math>, <math>DC = n</math> and <math>AD = d</math>, we have that <math>b^2m + c^2n = amn + d^2a</math>. (This is also often written <math>man + dad = bmb + cnc</math>, 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. |
<center>[[Image:Stewart's_theorem.png]]</center> | <center>[[Image:Stewart's_theorem.png]]</center> | ||
− | == Proof == | + | == Proof 1 == |
Applying the [[Law of Cosines]] in triangle <math>\triangle ABD</math> at [[angle]] <math>\angle ADB</math> and in triangle <math>\triangle ACD</math> at angle <math>\angle CDA</math>, we get the equations | Applying the [[Law of Cosines]] in triangle <math>\triangle ABD</math> at [[angle]] <math>\angle ADB</math> and in triangle <math>\triangle ACD</math> at angle <math>\angle CDA</math>, we get the equations | ||
*<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math> | *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math> | ||
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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>man + | + | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. |
+ | |||
+ | == Proof 2 (Pythagorean Theorem) == | ||
+ | |||
+ | Let the [[altitude]] from <math>A</math> to <math>BC</math> meet <math>BC</math> at <math>H</math>. Let <math>AH=h</math>, <math>CH=x</math>, and <math>HD=y</math>. So, applying [[Pythagorean Theorem]] on <math>\triangle AHC</math> yields | ||
+ | |||
+ | <cmath>b^2m = m(h^2+(y+n)^2) = m(h^2+y^2+2yn+n^2).</cmath> | ||
+ | |||
+ | Since <math>m=x+y</math>, <cmath>b^2m = m(h^2+y^2+2yn+n^2) = (x+y)(h^2+y^2+2yn+n^2) = h^2x+y^2x+2ynx+x^2+yh^2+y^3+2y^2n+n^2y.</cmath> | ||
+ | |||
+ | Applying Pythagorean on <math>\triangle AHD</math> yields | ||
+ | |||
+ | <cmath>c^2n = n(x^2+h^2) = nx^2+nh^2.</cmath> | ||
+ | |||
+ | Substituting <math>a=x+y+n</math>, <math>m=x+y</math>, and <math>d^2=h^2+y^2</math> in <math>amn</math> and <math>d^2a</math> gives | ||
+ | |||
+ | <cmath>amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}</cmath> | ||
+ | <cmath>d^2a = (h^2+y^2)(x+y+n) = h^2x+h^2y+h^2n+y^2x+y^3+y^2n.</cmath> | ||
+ | |||
+ | Notice that | ||
+ | |||
+ | <cmath>b^2m+c^2n = h^2+y^2x+2ynx+xn^2+yh^2+y^3+2y^2n+n^2y+nx^2+nh^2 \text{ and}</cmath> | ||
+ | <cmath>amn+d^2a = x^2n+2xyn+xn^2+y^2n+n^2y+h^2x+h^2y+h^2n+y^2x+y^3+y^2n</cmath> | ||
+ | are equal to each other. Thus, <math>b^2m + c^2n = amn+d^2a.</math> Rearranging the equation gives Stewart's Theorem: | ||
+ | |||
+ | <cmath>man+dad = bmb+cnc</cmath> | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
+ | ==Proof 3 (Barycentrics)== | ||
+ | Let the following points have the following coordinates: | ||
+ | |||
+ | <math>A: (1,0,0)</math> | ||
+ | |||
+ | <math>B: (0,1,0)</math> | ||
+ | |||
+ | <math>C: (0,0,1)</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>\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 | ||
==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ||
https://youtu.be/jEVMgWKQIW8 | https://youtu.be/jEVMgWKQIW8 | ||
− | ~ | + | ~IceMatrix |
== See also == | == See also == |
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