Difference between revisions of "Stewart's theorem"
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<cmath>amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}</cmath> | <cmath>amn = n(x+y+n)(x+y) = x^2n+2xyn+xn^2+y^2n+n^2y \text{ and}</cmath> | ||
− | <cmath>d^ | + | <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 | Notice that |
Revision as of 15:32, 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:
~sml1809
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix