Harriot, Thomas, Mss. 6784

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481241
[Commentary:
There is a reference on this page to Proposition II.12 from Conicorum libri quattuor (Apollonius . ]
De
[Translation: On ]
Sit datus angulus. xcd.
et punctum. a.
oportet ducere lineam aef
ut sit
ef=mn.
[Translation: Let there be a given angle xcd and a point a.
It is required to draw a line aef so that ef=mn, a given line.
fiat: ab=dc.
xc producta, in b.
fiat: ad=bc.
per punctum d, ad angulum abx
fiat hyperbola: dg.
Inscribatur dg=dk=mn.
fiat gf=dc
agatur af, quæ secabit dc, in e.
Dico quod: ef=mn,
[Translation: Make ab=dc, meeting with xc extended at b.
Make ad=bc.
Through the point d, at angle abx, make the hyperbola dg.
Inscribe dg=dk=mn.
Make gf=dc
Construct af, which will cut dc at e.
I say that ef=mn, the given line.
per. 12,2i
[Translation: by II.12 of the ]
Aliter.
Et continuetur ad utrasque partes.
usque ad p, q.
fiat: af=qp, quæ secabit dc in e.
Dico quod: ef=dg
[Translation: Another way.
And continuing on both sides, as far as p and q, make af=qp, which will cut dc at e.
I say that ef=dg.
Quoniam, ad=fp
erit: af=dp=gq. propter asymptotes
sed: ae=qd. ob parallelismum, cd, bg
Ergo: ef=dg
[Translation: Since ad=fp, therefore af=dp=gq on account of the asymptotes.
But ae=qd because cd and bg are parallel.
Therefore ef=dg.

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