Difference between revisions of "Stewart's theorem"
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<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} - 2nd\cos{\angle CDA} = b^{2} </math> | *<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math> | ||
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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. | ||
+ | |||
+ | == 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^a = (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> | ||
==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== |
Revision as of 02:57, 10 March 2022
Contents
[hide]Statement
Given a triangle with sides of length opposite vertices are , , , respectively. If cevian is drawn so that , and , we have that . (This is also often written , a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
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:
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix