Harriot, Thomas, Mss. 6784

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[Commentary:
On this page Harriot investigates Proposition 15 from Supplementum geometriæ (Viète 1593c, Prop .
Proposition XV.
Si e circumferential circuli cadant in diametrum perpendiculares duæ, una in centro, altera extra & ad perpendicularem in centro agatur ex puncto incidentiæ perpendicularis alterius, linea recta faciens cum diametro angulum æqualem trienti recti; a puncto autem quo acta illa secat perpendiculare in centro, ducatur alia linea recta ad angulum semicirculi: triplum quadratum huius, æquale est tam quadrato perpendicularis quae incidit extra centrum, quam quadratis segmentorum diametri, inter quæ perpendicularis illa media est

If from the circumference of a circle there fall two perpendiculars onto the diameter, one to the centre, the other off-centre; and to the perpendicular to the centre there is drawn from the point of incidence of the other perpendicular a straight line making an angle equal to one-third of a right angle to the diameter; moreover from the point where that line cuts the perpendicular to the centre, there is drawn another line to the angle of the semicircle, then three times the square of it is equal to the square of the perpendicular which falls off-centre and the squares of the segments of the diameter between which the perpendicular is the mean
The working contains a reference to Euclid's Elements, Proposition .
If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the parts, together with twice the rectangle contained by the parts. ]
prop. 15.
[Translation: Proposition 15 from the ]
Si e circumferential circuli cadant in
diametrum perpendiculares duæ; una in
centro; altera extra centrum: et ad per-
pendicularem in centro agatur ex puncto
incidentiæ perpendicularis alterius, linea
recta faciens cum diametro angulum æqualem
trienti recti, a puncto autem quo acta illa secat
perpendiculare in centro, ducatur alia
linea recta ad angulum semicirculi; Triplum
quadratum huius, æquale est tam quadrato perpendicularis quae incidit extra centrum,
quam quadratis segmentorum diametri, inter quæ perpendicularis illa media est

[Translation: If from the circumference of a circle there fall two perpendiculars onto the diameter, one to the centre, the other off-centre; and to the perpendicular to the centre there is drawn from the point of incidence of the other perpendicular a straight line making an angle equal to one-third of a right angle to the diameter; moreover from the point where that line cuts the perpendicular to the centre, there is drawn another line to the angle of the semicircle, then three times the square of it is equal to the square of the perpendicular which falls off-centre and the squares of the segments of the diameter between which the perpendicular is the mean ]

Sit diameter circuli ABC, a cuius circumferentia cadat perpendiculariter DB et fit
AB minus segmentum, BC maius, E verum centro. Sed et cadat quoque e circumferentia
perpendiculariter FE, et ex B ducatur recta BG ita ut angulus GBE sit æqualis trienti
recti, unde fiat BG dupla ipsius GE; et iungatur AG. Dico triplum quadratum ex AG
æquari quadrato ex DB, una cum quadrato ex AB et quadrato ex BC
[Translation: Let ABC be the diameter of a circle, from whose circumference there falls perpendicularly DB, and let AB be the lesser segment, BC the greater, and E the centre. But there also falls perpendicularly from the circumference FE, and from B there is drawn a line BG so that the angle GBE is equal to a third of a right angle, whence BG is twice GE; and AG is joined. I say that three times the square on AG is equal to the square on DB together with the square on AB and the squareon BC.

Etiam
per 4,2 El.
[…] Addatur utrovisque
[…] Ergo

[Translation: Also by Elements II.4

Hence the ]
Hinc tale Consectarium potest
[Translation: Here a Consequence of this kind may be ]
Datis tribus continue proportionalibus: invenire lineam cuius
quadratum sit tertia pars adgregati quadratorum e tribus

[Translation: Given three continued proportionals, find a line whose square is a third of the sum of the squares of all three proportionals.

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