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[Commentary:
The reference at the top of this page is to Zetetic 9 from Book III of Zeteticorum libri quinque
(Viète 1593a, Zet .
Zeteticum IX
Invenitur triangulum rectangulum numero.
Enim vero,
Adsumptis duobus lateribus rationalibus, hypotenusa fit similis adgregata quadratorum, basis differentia corumdem, perpendiculum duplo sub lateribus rectangulo.
Sint duo latero B & D. Sunt igitur proportionalia tria latera B, D, D quadratum/B. Omnia in B. Sunt tria proprtionalia Bq. Bin D. Dq. A quibus proportionalibus fit per antecdicta, hypotenusa trianguli similis Bq + Dq. basis Bq=Dq. perpendiculum B in D2. Et alioqui jam ordinatum est. Quadratum ab adgregato quadratorum, aequare quadratum a differentia quadratorum, adjunctum quadrato dupli rectanguli sub lateribus.
Sit B2. D3. Hypotenusa fiet similis 13, basis 5, perpendiculum
To find a right-angled triangle in numbers.
Taking two rational sides, the hypotenuse is similar to the sum of the squares, the base to their difference, the perpendicular to twice the product.
Let the two sides be and . There are therefore three proportionals , , . [Multiply] all by . There are three proportionals , , . From which proportionals it comes about, from what has been said before [see Zeteticum III.8], that the hypotenuse of the triangle is similar to , the base to , the perpendicular to . And now the rest is in order. The square of the sum of squares is equal to the square of the difference of squares added to the square of twice the product.
Suppose , . The hypotenuse is similar to 13, the base to 5, the perpendicular to 12.
Harriot followed the same instructions, replacing Viète's , , by , , reserving the letter for the base of the triangle. He denotes the quantities , , by (hypotenuse), (perpendicular), and (base), respectively, and demonstrates that , as required.
Note his use of what looks like an = sign in the first appearance of . This indicates that the positive difference is to be taken if . In modern notation, Harriot's would be written . ]
Zeteticum IX
Invenitur triangulum rectangulum numero.
Enim vero,
Adsumptis duobus lateribus rationalibus, hypotenusa fit similis adgregata quadratorum, basis differentia corumdem, perpendiculum duplo sub lateribus rectangulo.
Sint duo latero B & D. Sunt igitur proportionalia tria latera B, D, D quadratum/B. Omnia in B. Sunt tria proprtionalia Bq. Bin D. Dq. A quibus proportionalibus fit per antecdicta, hypotenusa trianguli similis Bq + Dq. basis Bq=Dq. perpendiculum B in D2. Et alioqui jam ordinatum est. Quadratum ab adgregato quadratorum, aequare quadratum a differentia quadratorum, adjunctum quadrato dupli rectanguli sub lateribus.
Sit B2. D3. Hypotenusa fiet similis 13, basis 5, perpendiculum
To find a right-angled triangle in numbers.
Taking two rational sides, the hypotenuse is similar to the sum of the squares, the base to their difference, the perpendicular to twice the product.
Let the two sides be and . There are therefore three proportionals , , . [Multiply] all by . There are three proportionals , , . From which proportionals it comes about, from what has been said before [see Zeteticum III.8], that the hypotenuse of the triangle is similar to , the base to , the perpendicular to . And now the rest is in order. The square of the sum of squares is equal to the square of the difference of squares added to the square of twice the product.
Suppose , . The hypotenuse is similar to 13, the base to 5, the perpendicular to 12.
Harriot followed the same instructions, replacing Viète's , , by , , reserving the letter for the base of the triangle. He denotes the quantities , , by (hypotenuse), (perpendicular), and (base), respectively, and demonstrates that , as required.
Note his use of what looks like an = sign in the first appearance of . This indicates that the positive difference is to be taken if . In modern notation, Harriot's would be written . ]
1.) Zet. lib. 3.
[Translation: Zetetica, Book III, Zeteticum ]
[Translation: Zetetica, Book III, Zeteticum ]