Stewart's theorem: Difference between revisions

Revision as of 20:14, 17 February 2025

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ and opposite vertices $A$ , $B$ , $C$ , respectively, then 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 $HD=y$ .

We can apply the Pythagorean Theorem on $\triangle AHC$ and $\triangle AHD$ to yield

$h^2 = b^2 - x^2 = d^2 - y^2$ and then solve for b to get $b^2 = d^2 + x^2 - y^2$

Doing the same for $\triangle AHB$ and $\triangle AHD$

$h^2 = c^2 - (m + y)^2 = d^2 - y^2$ then solve for c to get $c^2 = d^2 + m^2 + 2my$

Now multiple the first expression by m and the second by n

$mb^2 = md^2 + m(x^2 - y^2)$

$nc^2 = nd^2 + m^2n + 2mny$

Next add these two expressions

$mb^2 + nc^2 = md^2 + m(x^2 - y^2) + nd^2 + m^2n + 2mny$ .

Then simpify as follows (we reapply x + y = n a few times while factoring)

$mb^2 + nc^2 = (m + n)d^2 + m(x+y)(x - y) + mn(n + 2y)$ .

$mb^2 + nc^2 = (m + n)d^2 + mn(x - y) + mn(n + 2y)$ .

$mb^2 + nc^2 = (m + n)d^2 + mn(x + y + n)$ .

$mb^2 + nc^2 = (m + n)d^2 + mn(m + n)$ .

$mb^2 + nc^2 = (m + n)(d^2 + mn)$ .

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 23: / Line 23: @@
 == 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
+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>.
-<cmath>b^2m = m(h^2+(y+n)^2) = m(h^2+y^2+2yn+n^2).</cmath>
+We can apply the [[Pythagorean Theorem]] on <math>\triangle AHC</math> and <math>\triangle AHD</math> to yield
-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>
+<math>h^2 = b^2 - x^2 = d^2 - y^2</math>  and then solve for b to get  <math>b^2 = d^2 + x^2 - y^2</math>
-Applying Pythagorean on <math>\triangle AHD</math> yields
+Doing the same for  <math>\triangle AHB</math> and <math>\triangle AHD</math>
-<cmath>c^2n = n(x^2+h^2) = nx^2+nh^2.</cmath>
+<math>h^2 = c^2 - (m + y)^2 = d^2 - y^2</math> then solve for c to get <math>c^2 = d^2 + m^2 + 2my</math>
-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
+Now multiple the first expression by m and the second by n
-<cmath>amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}</cmath>
+<math>mb^2 = md^2 + m(x^2 - y^2)</math>
-<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
+<math>nc^2 = nd^2 + m^2n + 2mny</math>
-<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>
+Next add these two expressions
-<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>
+<math>mb^2 + nc^2 = md^2 + m(x^2 - y^2) + nd^2 + m^2n + 2mny</math>.
+Then simpify as follows (we reapply x + y = n a few times while factoring)
+<math>mb^2 + nc^2 = (m + n)d^2 + m(x+y)(x - y) + mn(n + 2y)</math>.
+<math>mb^2 + nc^2 = (m + n)d^2 + mn(x - y) + mn(n + 2y)</math>.
+<math>mb^2 + nc^2 = (m + n)d^2 + mn(x + y + n)</math>.
+<math>mb^2 + nc^2 = (m + n)d^2 + mn(m + n)</math>.
+<math>mb^2 + nc^2 = (m + n)(d^2 + mn)</math>.
+Rearranging the equation gives Stewart's Theorem:
+<math>man+dad = bmb+cnc</math>
 ~sml1809

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