Difference between revisions of "Stewart's theorem"

Revision as of 12:56, 16 August 2022

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 form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")

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

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> and 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 form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
 <center>[[Image:Stewart's_theorem.png]]</center>

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