Stewart's Theorem In triangle ABC, if D lies on the line BC and the segments have lengths AB=c, BC=a, CA=b, AD=d, BD=m, DC=n, then (1) if D lies between B and C, then a(d^{2}+mn) = mb^{2}+nc^{2}, (2) if D lies beyond C, then a(d^{2}mn) = mb^{2}nc^{2}, (3) if D lies beyond B, then a(d^{2}mn) =mb^{2}+nc^{2}.


Proof By the Cosine Rule applied to triangles ABD, ADC, we have cos(<ADB) = (d^{2}+m^{2}c^{2})/2dm, cos(<ADC) = (d^{2}+n^{2}b^{2})/2dn.
(1) Here <ADC = π<ADB, so cos(<ADC)=cos(<ADB), so that
(2) Now <ADC = <ADB, so we get
(3) The algebra is identical
to that in (2), but here nm=a, so

case (1) case (2) 