hyperbolic geometry

proof of stewart's theorem in euclidean geometry

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(d²+mn) = mb²+nc²,
(2) if D lies beyond C, then a(d²-mn) = mb²-nc²,
(3) if D lies beyond B, then a(d²-mn) =-mb²+nc².

Proof
By the Cosine Rule applied to triangles ABD, ADC, we have
cos(<ADB) = (d²+m²-c²)/2dm, cos(<ADC) = (d²+n²-b²)/2dn.
(1) Here <ADC = π-<ADB, so cos(<ADC)=-cos(<ADB), so that
(d²+n²-b²)/2dn = -(d²+m²-c²)/2dm. Simplifying this, we get
(m+n)(d²+mn) = mb²+nc². As m+n=a, we have (1).
(2) Now <ADC = <ADB, so we get
(d²+n²-b²)/2dn = (d²+m²-c²)/2dm. Simplifying this, we get
(m+n)(d²-mn) = mb²-nc². As m-n=a, we have (2).
(3) The algebra is identical to that in (2), but here n-m=a, so
we get (3).

case (1)

case (2)

stewart's theorem

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(d²+mn) = mb²+nc², (2) if D lies beyond C, then a(d²-mn) = mb²-nc², (3) if D lies beyond B, then a(d²-mn) =-mb²+nc².
Proof By the Cosine Rule applied to triangles ABD, ADC, we have cos(<ADB) = (d²+m²-c²)/2dm, cos(<ADC) = (d²+n²-b²)/2dn. (1) Here <ADC = π-<ADB, so cos(<ADC)=-cos(<ADB), so that (d²+n²-b²)/2dn = -(d²+m²-c²)/2dm. Simplifying this, we get (m+n)(d²+mn) = mb²+nc². As m+n=a, we have (1). (2) Now <ADC = <ADB, so we get (d²+n²-b²)/2dn = (d²+m²-c²)/2dm. Simplifying this, we get (m+n)(d²-mn) = mb²-nc². As m-n=a, we have (2). (3) The algebra is identical to that in (2), but here n-m=a, so we get (3).	case (1) case (2)