Difference between revisions of "Stewart's Theorem"

Revision as of 16:33, 18 February 2025

Statement

Given any 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.

Proof

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 following 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 $man + dad = bmb + cnc,$ as desired.

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

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

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

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

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 simplify 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:

$man+dad = bmb+cnc$

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: $\boxed{d^2a+amn=b^2m+c^2n}$

Video Proof

TheBeautyofMath

@@ Line 1: / Line 1: @@
-#REDIRECT[[Stewart's theorem]]
+== Statement ==
+Given any [[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, then 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.
+[[Image:Stewart's_theorem.png]]
+== 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 following equations:
+* <math>n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}</math>
+* <math>m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}</math>
+Because angles <math>\angle ADB</math> and <math>\angle CDA</math> are [[supplementary]], <math>m\angle ADB = 180^\circ - m\angle CDA</math>.  We can therefore solve both equations for the cosine term.  Using the [[trigonometric identity]] <math>\cos{\theta} = -\cos{(180^\circ - \theta)}</math> gives us
+*<math>\frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
+*<math>\frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>
+Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>. However, <math>m+n = a</math> so
+<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>man + dad = bmb + cnc,</math> as desired.
+=== 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>.
+We can apply the [[Pythagorean Theorem]] on <math>\triangle AHC</math> and <math>\triangle AHD</math> to yield
+<cmath>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</cmath>
+Doing the same for <math>\triangle AHB</math> and <math>\triangle AHD</math>
+<cmath>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</cmath>
+Now multiple the first expression by m and the second by n
+<cmath>mb^2 = md^2 + m(x^2 - y^2)</cmath>
+<cmath>nc^2 = nd^2 + m^2n + 2mny</cmath>
+Next add these two expressions
+<math>mb^2 + nc^2 = md^2 + m(x^2 - y^2) + nd^2 + m^2n + 2mny</math>.
+Then simplify as follows (we reapply x + y = n a few times while factoring):
+<cmath>mb^2 + nc^2 = (m + n)d^2 + m(x+y)(x - y) + mn(n + 2y)</cmath>
+<cmath>mb^2 + nc^2 = (m + n)d^2 + mn(x - y) + mn(n + 2y)</cmath>
+<cmath>mb^2 + nc^2 = (m + n)d^2 + mn(x + y + n)</cmath>
+<cmath>mb^2 + nc^2 = (m + n)d^2 + mn(m + n)</cmath>
+<cmath>mb^2 + nc^2 = (m + n)(d^2 + mn)</cmath>
+Rearranging the equation gives Stewart's Theorem:
+<cmath>man+dad = bmb+cnc</cmath>
+== 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>
+== Video Proof ==
+[//youtu.be/jEVMgWKQIW8 TheBeautyofMath]
+== See Also ==
+* [[Menelaus' theorem]]
+* [[Ceva's theorem]]
+* [[Geometry]]
+* [[Angle Bisector Theorem]]
+[[Category:Geometry]]
+[[Category:Theorems]]
+{{stub}}

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