10
Ladd, The Pascal Hexagram.
These three 
lines meet in a point, namely, the Brianchom point 
hence the two triangles formed by the vertical rows of 
points are homologous. The sides of the first are the Pascals 
and the corresponding sides of the second are 
hence these three pairs of sides intersect in a line which, as it is an axis of homology corresponding to the centre of homology we shall call the line 
In the same way it may be shown that the triangle formed by any three of the four Pascals to which the triangle 
belongs are homologous therewith, therefore the intersections of the four axes of homology, 
with the four Pascal lines 
respectively, are four points on one straight line. As this line is obtained by means of the triangle we shall call it the line 
To each triangle formed by three fundamental lines, no two of which pass through the same point on the conic, corresponds a line 
of the same notation; there are 
such triangles, hence there are lines 
To each 
point corresponds a 
line, hence there are 
lines 
divided into groups of four each, which intersect corresponding 
lines on the lines.
4. The triangles 
are homologous. Let us call their centre of homology 
their axis 
Let us say that the points 
are joined by the line 
and that the lines 
intersect in 
where the bar is drawn over the letters that are repeated. I have shown (Educational Times, Question 5698,) that 
intersect in 
and that 
are connected by 
There are 
points 
and lines 
Each point is joined to 
other points by 
lines, hence there are 
lines 
which pass by twos through the 
points 
and 
points 
which lie in twos on the lines 
The six lines
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, |
 |
, |
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 |
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intersect in pairs in three points on one straight line, viz., the points on 
hence they form the sides of a Pascal hexagon; and for a similar reason the six points of the same notation are the vertices of a Brianchon hexagon.
5. The Brianchon hexagon formed by joining alternate vertices of 
has for its sides 
The conic inscribed
