stewart's theorem in hyperbolic geometry

In euclidean geometry, the length of a cevian of a triangle is related to
the length of the sides by

 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. The proof in each case consists of an application of the Cosine Rule. Note that, if we regard a, m and n as signed lengths, with BC as the positive direction, then the formula in (1) covers all cases. If we use signed ratios along the line BC, then we can give a unified version. In the notation of Stewart's Theorem, for any point D on the line BC, d2 = (BD/BC)b2+(CD/CB)c2-(BD/BC)(CD/CB)a2. For the last term on the right, we have written mn as (m/a)(n/a)a2. When AD is a median, the formula simplifies since |BD|=|DC|=|BC|/2. Then we have an easy proof of part (1) of the following Corollary (1) If AD is a median of ΔABC, then |AD|2 = (2|AB|2+2|AC|2-|BC|2)/4. (2) If the other medians are BE and CF, then |AD|2+|BE|2+|CF|2 = 3(|AB|2+|BC|2+|CA|2)/4. (3) If ΔABC is equilateral of side a, then each median has length (/3/2)a. Part (2) is obtained by applying part (1) to each median, and adding the results. Part (3) also follows from part (1) since, for an equilateral triangle, a = b = c. This gives d2= (3/4)a2, and the result follows. Of course, there are easier ways to prove this last part, but our method adapts easily to hyperbolic geometry. Most of these euclidean results have hyperbolic counterparts, and the proofs of these run in parallel to the euclidean proofs, as we now show. Stewart's Theorem in Hyperbolic Geomatry In hyperbolic triangle ABC, if D lies on the hyperbolic line BC and the segments have hyperbolic lengths d(A,B)=c, d(B,C)=a, d(C,A)=b, d(A,D)=d, d(B,D)=m, d(D,C)=n, then (1) if D lies between B and C, then sinh(a)cosh(d) = sinh(m)cosh(b)+sinh(n)cosh(c), (2) if D lies beyond C, then sinh(a)cosh(d) = sinh(m)cosh(b)-sinh(n)cosh(c), (3) if D lies beyond B, then sinh(a)cosh(d) =-sinh(m)cosh(b)+sinh(n)cosh(c). The proof is an analogue of the euclidean proof, with the hyperbolic Cosine Rule in place of the euclidean version. Once again, if we look at signed (hyperbolic) ratios, we get a unified result. In the notation of Stewart's Theorem, for any point D on the hyperbolic line BC, cosh(d) = -h(D,B,C)cosh(b)-h(D,C,B)cosh((c) Recall that h(X,Y,Z) = ±sinh(d(X,Y))/sinh(d(Y,Z)) with the positive sign if and only if Y lies between X and Z. Thus, if D lies between B and C, then we have h(D,B,C) =-sinh(d(D,B))/sinh(d(B,C) =-sinh(m)/sinh(a). The other cases are similar. As in the euclidean case, we get a simple result for the hyperbolic length of a median of a triangle. Corollary (1) If AD is a median of the hyperbolic triangle ABC, then cosh(d) = (cosh(b)+cosh(c))/2cosh(½a). (3) If ΔABC is a equilateral hyperbolic triangle of side a, then each median has length d where cosh(d) = cosh(a)/cosh(½a). This follows from case (1) since for the median AD, we have m = n = ½a, and, for any a, sinh(a) =2 sinh(½a)cosh(½a). Part (3) folows at once since here a = b = c. There is no obvious (neat) analogue of part (2) of the euclidean corollary. It is appropriate at this point to mention another result on eq uilateral hyperbolic triangles. We know that the angles of such a triangle must all be equal. In the euclidean case, each angle is π/3, so all equilateral euclidean triangles are similar. In hyperbolic geometry, triangles with eqaul angles are congruent, so we cannot expect an analogous result. In fact, the angles of an equilateral hyperbolic triangle depend on the (common) hyperbolic length of the sides. Theorem If ΔABC is an equilateral hyperbolic triangle with sides of hyperbolic length a, then each of its angles is equal to θ, where cos(θ) = cosh(a)/(cosh(a)+1). Proof Applying the hyperbolic Cosine Rule to the traingle, we get cos(θ) = (cosh(a)cosh(a)-cosh(a))/sinh(a)sinh(a) = cosh(a)(cosh(a)-1)/(cosh2(a)-1). Cancelling (cosh(a)-1) from top and bottom, we get the result. Note that this gives a proof that all the angles are equal, but it also finds the angle θ. Also, as cosh(a) > 1, cosh(a)/(cosh(a)+1) > ½, so θ < π/3. Of course, this can also be deduced from the fact that the sum of the angles is less than π, so that 3θ < π.