Difference between revisions of "Stewart's theorem"
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== Statement == | == Statement == | ||
− | 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 | + | 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. |
<center>[[Image:Stewart's_theorem.png]]</center> | <center>[[Image:Stewart's_theorem.png]]</center> | ||
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<cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | ||
This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. | ||
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== Proof 2 (Pythagorean Theorem) == | == Proof 2 (Pythagorean Theorem) == | ||
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<math>C: (0,0,1)</math> | <math>C: (0,0,1)</math> | ||
− | <math>D: (0, \frac{n}{m+n},\frac{m}{m+n})</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>(1, -\frac{n}{m+n}, -\frac{m}{m+n})</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> | + | 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> |
~kn07 | ~kn07 |
Latest revision as of 16:37, 17 December 2024
Contents
[hide]Statement
Given a triangle with sides of length and opposite vertices , , , respectively, then if cevian is drawn so that , and , we have that . (This is also often written , 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 at angle and in triangle at angle , we get the equations
Because angles and are supplementary, . We can therefore solve both equations for the cosine term. Using the trigonometric identity gives us
Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: . However, so and This simplifies our equation to yield or Stewart's theorem.
Proof 2 (Pythagorean Theorem)
Let the altitude from to meet at . Let , , and . So, applying Pythagorean Theorem on yields
Since ,
Applying Pythagorean on yields
Substituting , , and in and gives
Notice that
are equal to each other. Thus, Rearranging the equation gives Stewart's Theorem:
~sml1809
Proof 3 (Barycentrics)
Let the following points have the following coordinates:
Our displacement vector has coordinates . Plugging this into the barycentric distance formula, we obtain Multiplying by , we get . Substituting with , we find Stewart's Theorem:
~kn07
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix