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

PROBLEM IV.

HAving the Hypothenusa of a right-angled Triangle given, and a mean proportional between the sides to find the Tri∣angle. As if the hypothenusa be AC (Fig. 44.) and a mean proportional between the sides BD, to find the sides AB and BC.

SOLƲTION.

Make the given Hypothenusa = a, and the mean propor∣tional =b, and the perpendicular BC=x; the basis AB by the hypoth. will be 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; and multiplying by xx, 〈 math 〉〈 math 〉; and substracting b{powerof4}, 〈 math 〉〈 math 〉.

Therefore by the third case, 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction. If a be put for unity, AC will be also = aa, and by making as AC to CG (a to b) so AF to GH (b to a third) this third will be GH=bb. Assuming therefore OC=½aa=OB the radius of a semi-circle, and ha∣ving erected CD=bb=BE parallel to it, EO will be

〈 math 〉〈 math 〉, and consquently EC = 〈 math 〉〈 math 〉, and EA 〈 math 〉〈 math 〉, viz. the double value of the quantity xx. Therefore for the double value of x, you must extract the roots out of them, i. e. you must find the mean proporti∣onals AL and AM between unity AC and AI=EC on the one side, and AK=AE on the other; Altho' these last may be more compendiously had, and the triangle it self immedi∣ately constructed, if having found EC and EA, you draw CB and AB: For these will be those two last mean proportionals = = AL and AM; for by reason of the ▵ ▵ ABC, AEB, and BEC, BC is a mean proportional between AC and CE, and AB a mean proportional between the same AC and AE by the 8. Lib. 6. Eucl. which is Consect. 3. Schol. 2. Prop. 34. Lib. 1. Math. Enucl.