A history of Greek mathematics - Wilbourhall.org

THE COLLECTION. BOOK VIII 435
is left intact, and let it be required to find the diameter of
a circular section of the cylinder. We take any two points
A, B on the surface of the fragment and by means of these we
find five points on the surface all lying in one plane section,
This is done by taking n\e different radii
in general oblique.
and drawing pairs of circles with A, B as centres and with
each of the five radii successively. These pairs of circles with
equal radii, intersecting at points on the surface, determine
five points on the plane bisecting AB at right angles.
The five
points are then represented on any plane by triangulation.
Suppose the points are A, B, C, D, E and are such that
^ no two of the lines connecting the different pairs
are parallel.
cr
Q!
This case can be reduced to the construction of a conic through
the five points A, B, D, E, F where EF is parallel to AB.
This is shown in a subsequent lemma (chap. 16).
For, if EF be drawn through E parallel to A B, and if CD
meet AB in and EF in 0', we have, by the well-known
proposition about intersecting chords,
C0.0D:A0.0B = CO' . O'D
: EC
.
O'F,
whence O'F is known, and F is determined.
We have then (Prop. 13) to construct a conic through A, B,
D, E, F, where EF is parallel to AB.
Bisect AB, EF W at V, ; then VW produced both ways
is a diameter. Draw DR, the chord through D parallel
F f 2