see, must needes lyghte within the circle. So if
you putte the pointes to be A.D, or D.C, or A.C, other B.C, or
B.D, in any of these cases you see, that the line that is drawen
from the one pricke to the other dothe euermore run within the
edge of the circle, els canne it be no right line. How be it,
that a croked line, especially being more croked then the
portion of the circumference, maye bee drawen from pointe to
pointe withoute the circle. But the theoreme speaketh only of
right lines, and not of croked lines.

_The xlviij. Theoreme._

If a righte line passinge by the centre of a circle, doo
crosse an other right line within the same circle, passinge
beside the centre, if he deuide the saide line into twoo
equal partes, then doo they make all their angles righte.
And contrarie waies, if they make all their angles righte,
then doth the longer line cutte the shorter in twoo partes.

_Example._

[Illustration]

The circle is A.B.C.D, the line that passeth by the centre, is
A.E.C, the line that goeth beside the centre is D.B. Nowe saye
I, that the line A.E.C, dothe cutte that other line D.B. into
twoo iuste partes, and therefore all their four angles ar righte
angles. And contrarye wayes, bicause all their angles are righte
angles, therfore it muste be true, that the greater cutteth the
lesser into two equal partes, accordinge as the Theoreme would.

_The xlix. Theoreme._

If twoo right lines drawen in a circle doo crosse one an
other, and doo not passe by the centre, euery of them dothe
not deuide the other into equall partions.

_Example._

[Illustration]

The circle is A.B.C.D, and the centre is E, the one line A.C,
and the other is B.D, which two lines crosse one an other, but
yet they go not by the centre, wherefore accordinge to the
woordes of the theoreme, eche of theim doth cuytte the other
into equall portions. For as you may easily iudge, A.C. hath one
portion longer and an other shorter, and so like wise B.D.
Howbeit, it is not so to be vnderstand, but one of them may be
deuided into ij. euen parts, but bothe to bee cutte equally in
the middle, is not possible, onles both passe through the centre,
therfore much rather when bothe go beside the centre, it can not
be that eche of theym shoulde be iustely parted into ij. euen
partes.

_The L. Theoreme._

If two circles crosse and cut one an other, then haue not
they both one centre.

_Example._

[Illust