Difference between revisions of "Stewart's theorem"
Etmetalakret (talk | contribs) (Created page with "== Statement == Given a triangle <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite vertices are <math>A</math>, <math>B</math>, <m...") |
|||
(20 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
== Statement == | == Statement == | ||
− | Given a [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> opposite [[vertex | vertices]] | + | 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. 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> | ||
− | == 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 equations | 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 | ||
− | *<math> n^{2} + d^{2} - | + | *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math> |
− | *<math> m^{2} + d^{2} - | + | *<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 | 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 | ||
Line 19: | Line 19: | ||
<cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and | <cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and | ||
<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>c^2n + b^2m = amn + d^2a,</math> | + | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. |
+ | |||
+ | == 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 | ||
+ | |||
+ | <cmath>b^2m = m(h^2+(y+n)^2) = m(h^2+y^2+2yn+n^2).</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Applying Pythagorean on <math>\triangle AHD</math> yields | ||
+ | |||
+ | <cmath>c^2n = n(x^2+h^2) = nx^2+nh^2.</cmath> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <cmath>amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}</cmath> | ||
+ | <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 | ||
+ | |||
+ | <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> | ||
+ | <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> | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ~kn07 | ||
+ | |||
+ | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ||
+ | https://youtu.be/jEVMgWKQIW8 | ||
+ | |||
+ | ~IceMatrix | ||
== See also == | == See also == |
Revision as of 11:03, 25 July 2024
Contents
[hide]Statement
Given a triangle with sides of length and opposite vertices , , , respectively. 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