the triangle to the right its edge B
will be brought into position to mark line _b_, also passing through
the centre of the circle. All that remains is to join the ends of these
two lines, using the square blade for lines _c_, _d_, and the triangle
for _e_ and _f_, its position on the square blade being denoted at T 3;
lines _c_, _d_, _e_, _f_, are the ones inked in.

[Illustration: Fig. 170.]

For a hexagon head we have the processes, Figures 170 and 171. The
circle is struck, and across it line _a_, Figure 170, passing through
its centre, the triangle of sixty degrees will mark the sides _b_, _c_,
and _d_, _e_, as shown, and the square blade is used for _f_, _g_.

[Illustration: Fig. 171.]

The chamfer circles are left out of these figures to reduce the number
of lines and so keep the engraving clear. Figure 171 shows the method of
drawing a hexagon head when the diameter across corners is given, the
lines being drawn in the alphabetical order marked, and the triangle
used as will now be understood.

[Illustration: Fig. 172.]

[Illustration: Fig. 173.]

It may now be pointed out that the triangle may be used to divide
circles much more quickly than they could be divided by stepping around
them with compasses. Suppose, for example, that we require to divide a
circle into eight equal parts, and we may do so as in Figure 172, line
_a_ being marked from the square, and lines _b_, _c_ and _d_ from the
triangle of forty-five degrees; the lines to be inked in to form an
octagon need not be pencilled, as their location is clearly defined,
being lines joining the ends of the lines crossing the circle, as for
example, lines _e_, _f_.

Let it be required to draw a polygon having twelve equal sides, and the
triangle of sixty is used, marking all the lines within the circle in
Figure 173, except _a_, for which the square blade is used; the only
lines to be inked in are such as _b_, _c_. In this example there is a
corner at the top and bottom, but suppose it were required that a flat
should fall there instead of a corner; then all we have to do is to set
the square blade S at the required angle, as in Figure 174, and then
proceed as before, bearing in mind that the point of the circle nearest
to the square blade, straight-edge, or whatever the triangle is rested
on, is always a corner of a polygon having twelve sides.

[Illustration: Fig. 174.]

[Illustration: Fig. 175.]

In both of these examples we have assumed that