Stewart's Theorem

Revision as of 08:51, 5 November 2006 by Bpms (talk | contribs) (CLarifying something)

Statement

Stewart's theorem.png

If a cevian of length d is drawn and divides side a into segments m and n, then

$cnc + bmb = man + dad.$


Proof

For this proof, we will use the law of cosines and the identity $\cos{\theta} = -\cos{(180 - \theta)}$.

Label the triangle $ABC$ with a cevian extending from $A$ onto $BC$, label that point $D$. Let CA = n. Let DB = m. Let AD = d. We can write two equations:

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

When we write everything in terms of cos(CDA) we have:

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

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n$. However, since $m+n$ can be written as a, we get the more common form: $cnc+bmb=man+dad$

See also