THE COLLECTION. BOOK YI1I 
435 
is left intact, and let it be required to tind 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, 
in general oblique. This is done by taking five different radii 
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 sA right angles. The five 
points are then represented on any plane by triangulation. 
Suppose the points are A, B, G, D, E and are such that 
no two of tlie lines connecting the different pairs are parallel. 
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 AB, and if CD 
meet AB in 0 and EF in O', we have, by the well-known 
proposition about intersecting chords, 
CO .OD-.AO .OB = GO'. O'l); EO'. 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 at V, W; then VW produced both ways 
is a diameter. Draw DR, the ghord through D parallel 
f f 2