Difference between revisions of "Stewart's theorem"

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.

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

@@ Line 1: / Line 1: @@
 == Statement ==
-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.")
+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>
-== 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
 *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math>
@@ Line 19: / Line 19: @@
 <cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and
 <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath>
-This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</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>
+<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==

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