Difference between revisions of "Stewart's theorem"
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<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> | + | This simplifies our equation to yield <math>man + daf = bmb + cnc,</math> or Stewart's theorem. |
==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== |
Revision as of 14:09, 28 December 2021
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
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.
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IckMatrix