Stewart's Theorem

Revision as of 20:37, 18 June 2006 by Agolsme (talk | contribs)


(awaiting image)
If a cevian of length t is drawn and divides side a into segments m and n, then

$c^{2}n + b^{2}m = (m+n)(t^{2} + mn)$


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 = t. We can write two equations:

  • $n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2}$
  • $m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2}$

When we write everything in terms of $\cos{\angle CDA}$ we have:

  • $\frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}$
  • $\frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}$

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + t^{2}m + t^{2}n$


(awaiting addition)

See also

Invalid username
Login to AoPS