134 
on pascal’s theorem. 
[403 
which appears thus, viz. 
line be contains points bef , beg , bch , 
= bf.cf, bg .eg, bh .eh, 
= 35.26, 16.34, 24.15; 
that is, be is the Pascalian line of the hexagon 162435; and the like for the rest of 
the six lines. 
The twenty points abc, abf,...fgli are as follows, viz. omitting the two points 
abc, fgh, the remaining eighteen points are the Pascalian points (the intersections of 
pairs of lines each through two of the points 1, 2, 3, 4, 5, 6) which lie on the 
Pascalian lines be, ca, ab, gh, hffg respectively ; the point abc is the intersection of 
the Pascalian lines be, ca, ab, and the point fgh is the intersection of the Pascalian 
lines gh, hf fg, the points in question being two of the points P (Steiner’s twenty 
points, each the intersection of three Pascalian lines). 
We thus see that we have two triads of hexagons such that the Pascalian lines 
of each triad meet in a point, and that the two points so obtained, together with the 
eighteen points on the six Pascalian lines, form a system of twenty points lying four 
together on fifteen lines, and which points and lines are the projections of the points 
and lines of intersection of six planes; or, say simply that the figure is the projection 
of the figure of six planes. 
It is to be added, that if the planes are a, b, c, f, g, h, then the point of 
projection is any one of the four points which have the same polar plane in regard 
to the system of the planes a, b, c, and in regard to the system of the planes f g, h. 
The consideration of the solid figure affords a demonstration of the existence as well 
of the six Pascalian lines as of the two points each the intersection of three of 
these lines.