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]] | + | 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 form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.") |
<center>[[Image:Stewart's_theorem.png]]</center> | <center>[[Image:Stewart's_theorem.png]]</center> |
Revision as of 12:56, 16 August 2022
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 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