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| − | == Statement ==
| + | #REDIRECT [[Stewart's Theorem]] |
| − | 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, 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. I know, it's easy to memorize.
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| − | <center>[[Image:Stewart's_theorem.png]]</center>
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| − | == Proof 1 ==
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| − | 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
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| − | *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math>
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| − | *<math> m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2} </math>
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| − | 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
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| − | *<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
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| − | *<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>
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| − | 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>.
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| − | However,
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| − | <math>m+n = a</math> so
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| − | <cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and
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| − | <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath>
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| − | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem.
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| − | == Proof 2 (Pythagorean Theorem) ==
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| − | 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>.
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| − | We can apply the [[Pythagorean Theorem]] on <math>\triangle AHC</math> and <math>\triangle AHD</math> to yield
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| − | <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>
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| − | Doing the same for <math>\triangle AHB</math> and <math>\triangle AHD</math>
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| − | <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>
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| − | Now multiple the first expression by m and the second by n
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| − | <math>mb^2 = md^2 + m(x^2 - y^2)</math>
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| − | <math>nc^2 = nd^2 + m^2n + 2mny</math>
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| − | Next add these two expressions
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| − | <math>mb^2 + nc^2 = md^2 + m(x^2 - y^2) + nd^2 + m^2n + 2mny</math>.
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| − | Then simpify as follows (we reapply x + y = n a few times while factoring)
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| − | <math>mb^2 + nc^2 = (m + n)d^2 + m(x+y)(x - y) + mn(n + 2y)</math>.
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| − | <math>mb^2 + nc^2 = (m + n)d^2 + mn(x - y) + mn(n + 2y)</math>.
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| − | <math>mb^2 + nc^2 = (m + n)d^2 + mn(x + y + n)</math>.
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| − | <math>mb^2 + nc^2 = (m + n)d^2 + mn(m + n)</math>.
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| − | <math>mb^2 + nc^2 = (m + n)(d^2 + mn)</math>.
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| − | Rearranging the equation gives Stewart's Theorem:
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| − | <math>man+dad = bmb+cnc</math>
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| − | ~sml1809
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| − | ==Proof 3 (Barycentrics)==
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| − | Let the following points have the following coordinates:
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| − | <math>A: (1,0,0)</math>
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| − | <math>B: (0,1,0)</math>
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| − | <math>C: (0,0,1)</math>
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| − | <math>D: \left(0, \frac{n}{m+n},\frac{m}{m+n}\right)</math>
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| − | 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>
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| − | ~kn07
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| − | ==Nearly Identical Video Proof with an Example by TheBeautyofMath==
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| − | https://youtu.be/jEVMgWKQIW8
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| − | ~IceMatrix
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| − | == See also ==
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| − | * [[Menelaus' theorem]]
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| − | * [[Ceva's theorem]]
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| − | * [[Geometry]]
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| − | * [[Angle Bisector theorem]]
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| − | [[Category:Geometry]]
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| − | [[Category:Theorems]]
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