649 
of Edinburgh, Session 1879-80. 
But L QLA = R, therefore LQ falls above LRQ, however small 
the distance AB may be ; in other words, LQM is concave towards 
AB. 
Every straight line terminated by a pair of conjugate equidistants 
to a given straight line is bisected by the given straight line , and 
makes equal alternate angles with the equidistants , c fc. 
If AB (fig. 7) be the given straight line, XP and YQ the equi- 
distants, POQ the line terminated by the equidistants, then the 
proposition follows at once by observing that, if AP and BQ be 
perpendiculars to AB, then AOP - BOQ. 
The common perpendicular to two conjugate equidistants is the 
least distance between them , the oblique distances are greater according 
as the angle they make with the perpendicular is greater , and the 
length of an oblique can be increased without limit. 
It will be seen that conjugate equidistants are analogous to Eucli- 
dean parallels. The analogy may be carried much farther. 
If equal arcs of two conjugate equidistants be joined towards the 
same parts by two straight lines, the figure so formed may be called 
a hyperbolic parallelogram. 
A mixed triangle whose base is the arc of an equidistant, whose 
two remaining sides are straight lines, and whose vertex lies on the 
conjugate equidistant, may be called a hyperbolic triangle. The 
following propositions are then very easily proved. 
The sum of the three angles of a hyperbolic triangle is 2R. 
The opposite straight sides of a hyperbolic parallelogram are equal 
to one another ; its diagonals bisect one another in a point on the 
straight line to which the equidistants that form its curved sides 
belong ; and each diagonal divides it into two congruent hyperbolic 
triangles. 
A series of propositions analogous to those of Euclid, Book I., 35- 
41, may be proved very easily ; we have only to substitute hyperbolic 
parallelograms and triangles for ordinary parallelograms and triangles, 
and conjugate equidistants for parallels. In particular, we see (fig. 8) 
that 
Two hyperbolic triangles CAOB, DA OB, which have for common 
base the arc AOB of an equidistant (and consequently have their ver- 
tices on the conjugate equidistant) are equal in area, 
Hence follows at once that — - 
