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Difference between revisions of "Stewart's theorem"

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== Statement ==
 
== Statement ==
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Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] are <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 form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
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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>
  
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== Proof ==
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== 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 + daf = bmb + cnc,</math> or Stewart's theorem.
+
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>
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 +
<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
  
−
~IckMatrix
+
~IceMatrix
  
 
== See also ==  
 
== See also ==  

Latest revision as of 11:03, 25 July 2024

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ and opposite vertices $A$, $B$, $C$, respectively. If cevian $AD$ is drawn so that $BD = m$, $DC = n$ and $AD = d$, we have that $b^2m + c^2n = amn + d^2a$. (This is also often written $man + dad = bmb + cnc$, 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

Proof 1

Applying the Law of Cosines in triangle $\triangle ABD$ at angle $\angle ADB$ and in triangle $\triangle ACD$ at angle $\angle CDA$, we get the equations

  • $n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}$
  • $m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}$

Because angles $\angle ADB$ and $\angle CDA$ are supplementary, $m\angle ADB = 180^\circ - m\angle CDA$. We can therefore solve both equations for the cosine term. Using the trigonometric identity $\cos{\theta} = -\cos{(180^\circ - \theta)}$ gives us

  • $\frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}$
  • $\frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}$

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$. However, $m+n = a$ so \[m^2n + n^2m = (m + n)mn = amn\] and \[d^2m + d^2n = d^2(m + n) = d^2a.\] This simplifies our equation to yield $man + dad = bmb + cnc,$ or Stewart's theorem.

Proof 2 (Pythagorean Theorem)

Let the altitude from $A$ to $BC$ meet $BC$ at $H$. Let $AH=h$, $CH=x$, and $HD=y$. So, applying Pythagorean Theorem on $\triangle AHC$ yields

\[b^2m = m(h^2+(y+n)^2) = m(h^2+y^2+2yn+n^2).\]

Since $m=x+y$, \[b^2m = m(h^2+y^2+2yn+n^2) = (x+y)(h^2+y^2+2yn+n^2) = h^2x+y^2x+2ynx+x^2+yh^2+y^3+2y^2n+n^2y.\]

Applying Pythagorean on $\triangle AHD$ yields

\[c^2n = n(x^2+h^2) = nx^2+nh^2.\]

Substituting $a=x+y+n$, $m=x+y$, and $d^2=h^2+y^2$ in $amn$ and $d^2a$ gives

\[amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}\] \[d^2a = (h^2+y^2)(x+y+n) = h^2x+h^2y+h^2n+y^2x+y^3+y^2n.\]

Notice that

\[b^2m+c^2n = h^2+y^2x+2ynx+xn^2+yh^2+y^3+2y^2n+n^2y+nx^2+nh^2 \text{ and}\] \[amn+d^2a = x^2n+2xyn+xn^2+y^2n+n^2y+h^2x+h^2y+h^2n+y^2x+y^3+y^2n\] are equal to each other. Thus, $b^2m + c^2n = amn+d^2a.$ Rearranging the equation gives Stewart's Theorem:

\[man+dad = bmb+cnc\]

~sml1809

Proof 3 (Barycentrics)

Let the following points have the following coordinates:

$A: (1,0,0)$

$B: (0,1,0)$

$C: (0,0,1)$

$D: \left(0, \frac{n}{m+n},\frac{m}{m+n}\right)$

Our displacement vector $\overrightarrow{AD}$ has coordinates $\left(1, -\frac{n}{m+n}, -\frac{m}{m+n}\right)$. Plugging this into the barycentric distance formula, we obtain \[d^2=-(m+n)^2 \left(\frac{mn}{(m+n)^2} \right)-b^2 \left ( -\frac{m}{m+n} \right)-c^2 \left(-\frac{n}{m+n}\right)=-mn+\frac{b^2m+c^2n}{m+n}\] Multiplying by $m+n$, we get $d^2(m+n)+mn(m+n)=b^2m+c^2n$. Substituting $m+n$ with $a$, we find Stewart's Theorem: \[\boxed{d^2a+amn=b^2m+c^2n}\]

~kn07

Nearly Identical Video Proof with an Example by TheBeautyofMath

https://youtu.be/jEVMgWKQIW8

~IceMatrix

See also

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