Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Stewart's Theorem"

(→‎Proof)
m
Line 10: Line 10:
  
 
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
 
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}{2\ce{nd}} = \cos{\angle CDA}</math>
+
*<math> \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}</math>
−
*<math> \frac{c^2 - m^2 -d^2}{2\ce{md}} = \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>.
 
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>.

Revision as of 14:21, 28 October 2020

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices are $A$, $B$, $C$, respectively. If cevian $AD$ is drawn so that $BD = m$, $DC = n$ and $AD = d$, we have that $b^2m + c^2n = amn + d^2a$. (This is also often written $man + dad = bmb + cnc$, a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")

Stewart's theorem.png

Proof

Applying the Law of Cosines in triangle $\triangle ABD$ at angle $\angle ADB$ and in triangle $\triangle ACD$ at angle $\angle CDA$, we get the equations

  • $n^{2} + d^{2} - 2\ce{nd}\cos{\angle CDA} = b^{2}$
  • $m^{2} + d^{2} - 2\ce{md}\cos{\angle ADB} = c^{2}$

Because angles $\angle ADB$ and $\angle CDA$ are supplementary, $m\angle ADB = 180^\circ - m\angle CDA$. We can therefore solve both equations for the cosine term. Using the trigonometric identity $\cos{\theta} = -\cos{(180^\circ - \theta)}$ gives us

  • $\frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}$
  • $\frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}$

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$. However, $m+n = a$ so \[m^2n + n^2m = (m + n)mn = amn\] and \[d^2m + d^2n = d^2(m + n) = d^2a.\] This simplifies our equation to yield $c^2n + b^2m = amn + d^2a,$ or Stewart's Theorem.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Stewart%27s_Theorem&oldid=135989"