648 Proceedings of the Royal Society 
If we take the latter alternative, we get Euclidean or homaloidal 
space ; and, from the defining property by which we have character- 
ised it, we can prove Euclid’s parallel axiom, and develop Euclid’s 
geometry in his or any other equivalent manner. 
Having separated out homaloidal space, let us now consider mere 
closely hyperbolic space proper, in which the defect is always positive. 
The fundamental proposition to be proved is the following. 
Tlie defect of a triangle ( and consequently the defect of any plane 
rectilineal figure) is proportional to its area. 
Various proofs of this proposition might be given. I select that 
which depends on the properties of the curves of equidistance from 
a straight line, because the intermediate propositions are the analogues 
in hyperbolic space to the propositions regarding parallels and 
parallelograms that are given in the latter part of Euclid’s first book. 
If in any plane perpendiculars of constant length be erected upon 
a given straight line, their extremities generate two curves which I 
shall call the equidistants, the two equidistants corresponding to a 
given length of the perpendicular may be called conjugate equidis- 
tants. 
The equidistant is a self congruent line. 
For if we take any piece AB (fig. 4) of the given line, and LM the 
corresponding piece of the equidistant, and if also A'B' = AB and 
L'M' be corresponding points to A' and B', then, if we place A'B' 
on AB, L' and M' will coincide with L and M, and, if A'P' = AP, Q' 
will coincide with Q, and so on. Hence the piece L'M' is congruent 
with the piece LM. 
The equidistant is at every point at right angles to the generating 
perpendicular. 
This is at once evident by considering two equal pieces (fig. 5) 
LP and LQ of the equidistant on either side of L, and the corres- 
ponding points A and B on the straight line, so that OA = OB. 
We have LOAP=:LOBQ, hence Z.OLP= L OLQ, each = B. 
The equidistant in hyperbolic space is a curved line , concave towards 
the given line. 
Let LQM (fig. 6) be a piece of the equidistant, LM a straight line 
cutting the perpendicular through P, the middle point of AB, 
in R. Then LRPA = MRPB. Hence L PEL = L PRM = R, and 
the angles at P are each = R, therefore ALR < R. 
