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Difference between revisions of "Stewart's theorem"

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Revision as of 16:01, 19 February 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. 

Good Job! You mastered 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

TOTO SLOT

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