To drawe a circle about any triangle assigned.
Fyrste deuide two sides of the triangle equally in half and from
those ij. prickes erect two perpendiculars, which muste needes
meet in crosse, and that point of their meting is the centre of
the circle that must be drawen, therefore sette one foote of the
compasse in that pointe, and extend the other foote to one
corner of the triangle, and so make a circle, and it shall
touche all iij. corners of the triangle.
_Example._
[Illustration]
A.B.C. is the triangle, whose two sides A.C. and B.C. are
diuided into two equall partes in D. and E, settyng D. betwene
B. and C, and E. betwene A. and C. And from eche of those two
pointes is ther erected a perpendicular (as you se D.F, and
E.F.) which mete, and crosse in F, and stretche forth the other
foot of any corner of the triangle, and so make a circle, that
circle shal touch euery corner of the triangle, and shal enclose
the whole triangle, accordinge, as the conclusion willeth.
An other way to do the same.
And yet an other waye may you doo it, accordinge as you learned
in the seuententh conclusion, for if you call the three corners
of the triangle iij. prickes, and then (as you learned there) yf
you seeke out the centre to those three prickes, and so make it
a circle to include those thre prickes in his circumference, you
shall perceaue that the same circle shall iustelye include the
triangle proposed.
_Example._
[Illustration]
A.B.C. is the triangle, whose iij. corners I count to be iij.
pointes. Then (as the seuentene conclusion doth teache) I seeke
a common centre, on which I may make a circle, that shall
enclose those iij prickes. that centre as you se is D, for in D.
doth the right lines, that passe by the angles of the arche
lines, meete and crosse. And on that centre as you se, haue I
made a circle, which doth inclose the iij. angles of the
triangle, and consequentlye the triangle itselfe, as the
conclusion dydde intende.