Harriot, Thomas, Mss. 6784

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[Commentary:
There is a reference on this and the four following folios to page 212 of Commandino's edition of Mathematicae collectiones (Pappus . Page 212 contains Proposition 85, also denoted Lemma XI.
Problema V. Propos. LXXXV.
Semicirculo positione dato ABC, & dato puncto D, describere per D semicirculum, qualis est DEF, ita vt ducatur contingens BC, fiat AD ipsi BE æqualis.

Given a semicircle ABC and a point D, draw through D a semicircle DEF, so that when the tangent BC is drawn, AD is equal to BE. ]
Pappus. 212.
Hic habetur usus determinatæ
[Translation: Here is found the use of a determinate ]
Semicirculo positione dato ABC, & dato puncto D: Describere per D semicirculum,
qualis est DEF, ita ut ducatur contingens BC, fiat AD ipsi BE
[Translation: Given a semicircle ABC and a point D, draw through D a semicircle DEF, so that when the tangent BC is drawn, AD is equal to BE.
Constructio:
Centro C, intervallo CD, describitur periferia DIQ; Et inscribatur DI
æqualis AD. A puncto I ducatur IL perpendicularis ad AC.
Dividatur DC bisariam in puncto K. fiat AH æqualis AD. et LM æqualis KC.
Dividatur MH bisariam in N. et fiat DO æqualis MN vel HN. Sit PD
perpendicularis ad AC. et ducatur OP, cui fiat æqualis OG. Dico quod
G est centrum semicirculo quæsiti. qui describatur et sit DEF. a puncto C
ducatur contingens CE et producatur ad B. Dico quod BE æqualis
est AD
[Translation: Construction
With centre C and radius CD, there isdrawn the circumference DIQ. And there is inscribed DI equal to AD. From the point I there is drawn IL perpendicular to AC. The line DC is bisected at the point K. Make AH equal to AD and LM equal to KC. The line MH is bisected at N, and make DO euqal to MN or HN. Let PD be perpendicular to AC and OP is drawn, to which make OG equal. I say that G is the centre of the semicircle sought, which is drawn, and is DEF. From the point C there is drawn the tangent CE extended to B. I say that BE is equal to AD.

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