View source for Stewart's theorem

== 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.  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>

== 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
*<math> n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2} </math>
*<math> m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2} </math>

Because angles <math>\angle ADB</math> and <math>\angle CDA</math> are [[supplementary]], <math>m\angle ADB = 180^\circ - m\angle CDA</math>.  We can therefore solve both equations for the cosine term.  Using the [[trigonometric identity]] <math>\cos{\theta} = -\cos{(180^\circ - \theta)}</math> gives us
*<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>

*<math> \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}</math>

Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>.
However, 
<math>m+n = a</math> so 
<cmath>m^2n + n^2m = (m + n)mn = amn</cmath> and 
<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.

 Good Job! You mastered 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^2a = (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>

~sml1809

==Nearly Identical Video Proof with an Example by TheBeautyofMath==
https://youtu.be/jEVMgWKQIW8

~IceMatrix

== See also == 
* [[Menelaus' theorem]]
* [[Ceva's theorem]]
* [[Geometry]]
* [[Angle Bisector theorem]]

[[Category:Geometry]]

[[Category:Theorems]]