To parte a line assigned after suche a sorte, that the
square that is made of the whole line and one of his parts,
shal be equal to the squar that cometh of the other parte
alone.
First deuide your lyne into ij. equal parts, and of the length
of one part make a perpendicular to light at one end of your
line assigned. then adde a bias line, and make thereof a
triangle, this done if you take from this bias line the halfe
lengthe of your line appointed, which is the iuste length of
your perpendicular, that part of the bias line whiche dothe
remayne, is the greater portion of the deuision that you seke
for, therefore if you cut your line according to the lengthe of
it, then will the square of that greater portion be equall to
the square that is made of the whole line and his lesser
portion. And contrary wise, the square of the whole line and his
lesser parte, wyll be equall to the square of the greater parte.
[Illustration]
_Example._
A.B, is the lyne assigned. E. is the middle pricke of A.B, B.C.
is the plumb line or perpendicular, made of the halfe of A.B,
equall to A.E, other B.E, the byas line is C.A, from whiche I
cut a peece, that is C.D, equall to C.B, and accordyng to the
lengthe lo the peece that remaineth (whiche is D.A,) I doo
deuide the line A.B, at whiche diuision I set F. Now say I, that
this line A.B, (w^{ch} was assigned vnto me) is so diuided in
this point F, y^t y^e square of y^e hole line A.B, & of the one
portion (y^t is F.B, the lesser part) is equall to the square of
the other parte, whiche is F.A, and is the greater part of the
first line. The profe of this equalitie shall you learne by the
.xl. Theoreme.
[Transcriber's Note:
There are two ways to make this Example work:
--transpose E and F in the illustration, and change one
occurrence of E to F in the text ("at whiche diuision I
set..."), _or_:
--keep the illustration as printed, and transpose all other
occurrences of E and F in the text.]