In euclidean geometry, the length of a cevian of a triangle is related to
the length of the sides by
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
In the notation of Stewart's Theorem, for any point D on the line BC,
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.
Most of these euclidean results have hyperbolic counterparts, and the proofs of
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
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,
Recall that h(X,Y,Z) = ±sinh(d(X,Y))/sinh(d(Y,Z)) with the positive sign if and
As in the euclidean case, we get a simple result for the hyperbolic length of a
This follows from case (1) since for the median AD, we have m = n = 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
Note that this gives a proof that all the angles are equal, but it also finds the angle θ.