650 
Proceedings of the Royal Society 
The rectilinear triangles CAB , DAB on the same chord of an 
equidistant, whose vertices lie on the conjugcde equidistant, are equal 
in area and defect . 
N. B. — the defect is 2 L OAB in both cases. It is obvious that, if we 
join the middle points of the sides of any triangle, the extremities of 
its base lie on an equidistant to the line so drawn, and the vertex 
lies on the conjugate equidistant. Bearing this in mind, the pro- 
perties of equidistants enable us to establish the following proposi- 
tions : — 
We can always construct an isosceles triangle whose base is equal 
to one side of a given triangle , and ichose area and defect are the 
same as those of the given triangle. 
Given two triangles , we can ahuays transform one or other of them 
into another of equal area and defect which has one of its sides equal 
to one of the sides of the remaining triangle * 
Hence two triangles that have the same area must have the same 
defect , and conversely , for we can transform them into a pair of 
isosceles triangles on the same base without altering either area or 
defect. It is obvious that two such triangles must be congruent 
if they are equal in area, and hence they must be equal in defect ; 
and from what I have proved concerning the defect of composite 
figures, the converse follows with equal ease. 
Hence the area of a triangle is proportional to its defect. Hence, 
p being a certain linear constant, characteristic of a hyperbolic space, 
and A the area of a rectilineal triangle of defect 8, we have 
A = P 2 8. 
A great variety of very important conclusions can at once be drawn 
from this formula. I mention some of the most interesting. 
Since 8 = ^-, if p be infinite, then 8 = 0 for every triangle of 
P 2 
finite area ; in other words, homaloidal space is simply a hyperbolic 
space whose linear constant is infinite. This conclusion may be 
looked at from another, but mathematically equivalent, point of 
view. Let us imagine a hyperbolic space of given linear constant p. 
* I leave the reader to consider and settle for himself whether a simpler pro- 
position than the above could be established. In particular he should consider 
the following problem in hyperbolic geometry: — “To construct an isosceles 
triangle of given area on a given base. ” 
