D and E rayse vp vnto the lines AB and BC perpendicular lynes DF and EF. Now forasmuch as either of these angles BDF, and BEF is a right angle, a right line produ∣ced from the point D to the point E, shall deuide either of the said angles: and foras∣much as it falleth vppon the right lines DF and EF, it shall make the inward angles on one and the selfe same side, namely, the angles DEF
and
EDF lesse then two right angles. Wherefore (by the fift peticion) the lines
DF and
EF being produced shall concurre. Let them concurre in the point
F. And forasmuch as a certaine right line
DF deuideth a cer∣taine right lyne
AB into two equall partes and per∣pendicularly, therfore (by the corollary of the first of this booke) in the line
DF is the centre of the circle, & by the same reason the centre of the selfe same circle shalbe in the right line EF. Wherfore the centre of the circle wherof
ABC is a section, is in the point
F, which is commō to either of the lines
DF and
EF. Wherfore a section of a circle being geuē, namely, the section
ABC, there is described the circle of the same section: which was required to be done.
And by this last generall way, if there be geuen three pointes, set howsoeuer, so that they be not all three in one right line, a man may describe a circle which shall passe by all the said three pointes. For as in the example before put, if you suppose onely the 3. pointes A, B, C, to be geuen and not the circumference ABC to be drawen, yet follow∣ing the selfe ••ame order you did before, that is, draw a right line from A to B and an other from B to C and deuide the said right lines into two equall parts, in the points D and E, and erect the perpendicular lines DF and EF cutting the one the other in the point F, and draw a ••ight line from F to B: and making the centre the point F, and the space FB describe a circle, and it shall passe by the pointes A & C: which may be pro∣ued by drawing right lines from A to F, and from F to C. For forasmuch as the two si••es AD and DF of the triangle ADF are equall to the two sides BD and DF of the triangle BDF (for by supposition the line AD is equall to the line DB, and the lyne DF is common to them bot••) and the angle ADF is equall to the angle BDF (for they are both right angles) therfore (by the 4. of the first) the base AF is equall to the base BF. And by the same reason the line FC is equall to the line FB. Wherefore these three lines FA, FB and FC are equall the one to the other. Wherefore makyng the centre the point F and the space FB, it shall also passe by the pointes A and C. Which was required to be done. This proposition is very necessary for many things as you shal afterward see.
Campane putteth an other way, how to describe the
whole circle of a sectiō geuen. Suppose that the section be
AB. It is required to describe the whole circle of the same section. Draw in the section two lines at all aduen∣tures
AC and
BD: which deuide into two equall parts
AC in the point
E, and
BD in the pointe
F. Then from the two pointes of the deuisions draw within the secti∣on two perpendicular lines
EG and
FH which let cutte the one the other in the point
K. And the centre of the circle shall be in either of the said perpendicular lines by the corollary of the first of this booke. Wherfore the point
K is the centre of the cir∣cle: which was required to be done.
But if the lines EG & FH do not cut the one the o∣ther,
but make one right line as doth
GH in the secōd figure: which happeneth when the two lines
AC and
BD are equidistant. Then the line
GH, being applyed to either part of the circumference geuen, shall passe by the centre of the circle, by the selfe same Corollary. For the lines
EG and
FH cannot be equidistant. For then one and the self same circumference should haue two centres. Wherfore the line
HG being deuided in∣to