Harriot, Thomas, Mss. 6785

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1910
Iisdem positis: In linea ad sumatur quovis punctum o
et erigatur perpendicularis on quæ erit parallela lineaæ
ah et secabit lineam he in puncto n. agatur etiam recta
na, quæ secabit gt in puncto e. Manifestum est
[Translation: The same things being supposed, in the line ad there is taken any point o and there is constructed the perpendicular on which will be parallel to the line ah and will cut the line he in the point n. There is also constructed the line na, which will cut gt in the point e.
Dico primo quod ae est maior quam ao.
connectantur puncta d, t,
et fiat ap æqualis lineæ ns
sitque acta recta ps quæ
necessario secabit lineam dt in puncto r.
Quoniam et tz in puncto q. Quoniam
In triangulo adt, latera
ad et at sunt æqualia,
anguli etiam ad d et t
sunt etiam
[Translation: I say first that ae is greater than ao.
Connect the points d and t, and make ap equal to the line ns; let there also be constructed the line ps which will necessarily cut the line dt in the point r and tz in the point q. Because in the triangle adt, the sides ad and at are equal, and also the angles at d and t are also equal.
Sed angulus ad t nempe atd hoc est atr est
æqualis duobus angulis interioris trs et tsr. Duobus videlicet
interioris et oppositis triangulis trs. Ergo maior est trs anguli,
maior etiam angulo prd qui est ad verticem
anguli trs. Ergo, angulos adr hoc est pdr est
maior quam angulus prd. Ergo pr lineam est maior quam pd.
multo igitur maior pq quam pd. Sed linea pq, æqualis est ae
ergo ae est maior quam pd. Sed etiam pd est æqualis ao, cum
sunt latera subtendentia æqualis angulos æqualium triangulum
nempe, dsp et onp. Ergo ae est maior quam ao, quod primo
demonstrandum
[Translation: But the angle at t, namely atd, that is atr is equal to two interior angles trs et tsr, clearly the two interior and opposite angles in triangle trs. Therefore trs is the greater angle.
The angle prd is also greater, which is at the vertex of angle trs. Therefore, the angle adr, that is pdr, is greater than the angle prd. Therefore the line pr is greater than pd, and therefore pq is much greater than pd. But the line pq is equal to ae, therefore ae is greater than pd. But also pd is equal to ao, since they are sides subtending equal angles of equal triangles, namely, dsp et ona. Therefore ae is greater than ao, which was to be demonstrated first.
Dico secundo, ut at ad as, seu ut ag ad ah, ita ae ad an.
Hoc enim manifestum ex similitudine triangulorum ate et asu,
seu aeg et anh
[Translation: I say, second, that as at to as, or as ag to ah, so is ae to an.
For this is clear from the similarity of triangles ate and asu, or aeg and anh.

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