Harriot, Thomas, Mss. 6785

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[Commentary:
On this page Harriot investigates Propositions 20 and 21 from Supplementum geometriæ (Viète 1593c, Props 20, .
Proposition XX.
Constituere triangulum aæquicrurum, ut differntia inter basin & alterum e cruribus fit ad basin, sicut quadratum cruris ad quadratum compositae ex crure &

To construct an isosceles triangle so that the difference between the base and either of the legs is to the base as the square of a leg is to the square of the sum of a leg and the base.
Proposition XXI.
Si fuerit triangulum aequicrurum, fit autem differentia inter basin & alterum e cruribus ad basin, sicut quadratum cruris ad quadratum compositæ ex crure & base: quae a termino basis ducetur ad crus linea recta ipsi cruri æquale, secabit bisariam angulum ad basin.

If there is an isosceles triangle, and moreover the [ratio of the] difference between the base and either of the legs, to the base, is equal to the square of the leg to the square of the sum of a leg and the base, then a line drawn from the [end of the] base to the leg, equal [in length] to that leg, will bisect the angle at the
There is a reference to Propostion 19 of the Supplementum (see Add MS 6785 f. ). There are also references to Euclid's Propositions , , , and .
The angle at the centre of a circle is double the angle at the circumference, when they have the same part of the circumference for a base.
If from a point without a circle two straight lines be drawn to it, one of which is a tangent to the circle, and the other cuts it; the rectangle under the whole cutting line and the external segment is equal to the square of the tangent.
To find a fourth proportional to three given lines.
Equal parallograms which have one angle each equal have the sides about the equal angles reciprocally proportional. ]
prop. 20.
[Translation: Proposition 20 from the ]
Constituere triangulum aæquicrurum; ut differntia inter basin et alterum
e cruribus fit ad basin, sicut quadratum cruris ad quadratum compositæ
ex crure et
[Translation: To construct an isosceles triangle so that the difference between the base and either of the legs is to the base, as the square of a leg is to the square of the sum of a leg and the base.
per 19,p fiat
Et ponatur in circumferentia, DE=AB vel AC.
Et ducantur recta AE
Triangulum AED est quod
[Translation: By Proposition 19,
And in the circumference, put DE=AB or AC. and constructing the line AE, the triangle AED is as required.
prop.
[Translation: Proposition ]
Si fuerit triangulum aequicrurum: fit autem differentia inter basin et alterum
e cruribus ad basin; sicut quadratum cruris ad quadratum compositæ ex crure et base.
Quae a termino basis ducetur ad crus linea recta ipsi cruri æquale: secabit
bisariam angulum ad
[Translation: If there is an isosceles triangle, and moreover the [ratio of the] difference between the base and either of the legs, to the base, is equal to the square of the leg to the square of the sum of a leg and the base, then a line drawn from the [end of the] base to the leg, equal [in length] to that leg, will bisect the angle at the ]
AF secat bisariam angulum EAD.
Nam ex hypothesi.
[…]
Et per 36,3 el
Ergo. per 14, 6 el
[…]
Consequenter:
Et subducendo
Ergo: per 2,6: el: EC et FA sunt parallelæ
Et: Angulus ECD æqualis angulo FAD.
Sed. per 20,3: Angulus EAD est duplus anguli ECD Hoc est: FAD
Ergo angulus EAD sectis est bisariam a recta AF.
Quod erat
[Translation: AF bisects the angle EAD.
For from the hypothesis

And by Elements III.36
Therefore by Elements VI.14

Consequently:
And subtracting
Therefore, by Elements VI.2, EC and FA are parallel.
And angle ECD is equal to angle FAD.
But by Elements III.20, angle EAD is twice angle ECD, that is FAD.
Therefore angle EAD is cut in two by the line AF.
Which was to be ]

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