Difference between revisions of "Stewart's theorem"
m (→Statement) |
|||
Line 1: | Line 1: | ||
== Statement == | == Statement == | ||
− | 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 | + | 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, then 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> |
Revision as of 10:26, 15 December 2024
Contents
[hide]Statement
Given a triangle with sides of length
and opposite vertices
,
,
, respectively, then 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