Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

SOLƲTION.

Denomination. Make AE, half AB=a, HE=x a•• HC=y; therefore AH will be = a−x and HB=a+•• whence the Denomination of the squares AC and BC is ea•• had; viz. the one 〈 math 〉〈 math 〉, and the other, a••

[illustration]

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xx+yy, so that the sum of the squares is 2aa+2xx+

And the ▵ ACB will be = ay: And since the ▵ ▵ •• and AIC are similar, and the sides of the former FD and ••rbitrary, so that for FD we may put b and for FC, c; ••he sides of the latter are determined by the similitude of ▵▵ ABI and HDB, as being right-angled ones, and ha∣•• the common angle B; they will be obtain'd by making the Hypothen. BC to the Hypoth. AB so the base HB to ••ase BI, i. e.

〈 math 〉〈 math 〉, whence substracting BC=e, there remains CI = 〈 math 〉〈 math 〉.

For the first Equation, by virtue of the Problem as 〈 math 〉〈 math 〉 to ay. And the Rectangle of the ••mes is = to the Rectangle of the means, i. e. 〈 math 〉〈 math 〉.

For the other Equation, since as DF to FG so CI to AI 〈 math 〉〈 math 〉 the Rectangle of the extreams will again be = to the Re∣••gle of the means, i. e. 〈 math 〉〈 math 〉; multiplying both sides by 〈 math 〉〈 math 〉; ••h is the second Equation.

The Reduction of both Equations. 〈◊〉〈◊〉 first was 〈 math 〉〈 math 〉. ••refore dividing by 〈 math 〉〈 math 〉. substracting 〈 math 〉〈 math 〉. ••he latter Equation was 〈 math 〉〈 math 〉, i. e. ••ituting again the value ee, which was 〈 math 〉〈 math 〉; and by transposition, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉, or 〈 math 〉〈 math 〉,

or (putting 2 f for 〈 math 〉〈 math 〉) 〈 math 〉〈 math 〉.

Wherefore we have the value of xx twice expressed, but by quantities partly unknown, because y is found on both sides. Wherefore now we must make a new comparison of their values, whence you'l have this new

5. Third Equation, in which there is only one of the un∣known quantities: 〈 math 〉〈 math 〉; and adding on both sides both yy and aa, 〈 math 〉〈 math 〉; or dividing by 2, 〈 math 〉〈 math 〉; and transposing 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; which is the value of the quantity y in known terms.

But this value in one of the precedent Equations, viz. in this 〈 math 〉〈 math 〉, being substituted for y and its square for yy, will give 〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉.

6. The Geometrical Construction, which Schooten gives us p. 153. Having made the angle KAB (n. 2. Fig. 29.) e∣qual to the given one D, erect from A, AL perpendicular to KA, meeting the Perpendicular EM in L; and from the Centre L, at the interval of the given right line d, describe a Circle that shall cut KA and EL produced to K and M. Then assuming EN=KA, join MA, and from N draw NH pa∣rallel to it, which shall meet AB in H. Afterwards, having described from L, at the interval LA, the segment of a Circle ACB, draw from H, HC perpendicular to AB meeting the circumference in C, and join AC, CB.

NB. The reason of this elegant Construction, which the Author conceal'd, for the sake of Learners we will here shew. 1. Therefore, he reduc'd the last Equation (extracting the root, as well as it could bear, both of Numerator and Deno∣minator) to this: 〈 math 〉〈 math 〉 multipl. by 〈 math 〉〈 math 〉, so that after this way the Construction would be reduc'd to this proportion, as d+f to a so 〈 math 〉〈 math 〉. 2. He made the angle KAE = to the given one D, and the angle KAL a right one, so that having described the segment of a Circle from L the inscribed angle will also be made equal to the gi∣ven one, according to the 33. Lib. 3. Eucl. 3. By doing this, EL expresses the quantity f, since by reason of the simi∣larity of the ▵ ▵ KOA s. GFD, n. 1. and AEL (for the angles LAE and AKO are equal, because each makes a right one with the same third Angle KAO) you have as KO to OA so AE to EL i. e. 〈 math 〉〈 math 〉

4. Making now LM and LK = d you had EM=d+f, and AK = 〈 math 〉〈 math 〉 (for the □ AL is = to aa+ff, which being substracted from □ LK = 〈 math 〉〈 math 〉.)

5. Wherefore there now remains nothing to construct the last Equation above, but to make EN=AK, and to draw HN parallel to AM; for thus was the whole proportion as EM to EA so EN to EH 〈 math 〉〈 math 〉 to x. Q. e. f.

For the point H being determined, a perpendicular HC thence erected in the segment already described defines the Point C, which answers the Question.