PROB. II. 1 * 24 Dat. To find three straight lines such that the ratio of the first to the second is given ; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the sum of the squares of the first and third is given. Let AB be the first straight line, BC the second, and BD the third : and because the ratio of AB to BC is given, and that if a given straight line be taken from BC, the ratio of the remainder to BD is given; therefore * the excess of the first AB above a given straight line, has a given ratio to the third BD: let AE be that given straight line, therefore the remainder EB has a given ratio to BD: let BD be placed at right angles to EB, and join DE; then the triangle EBD is * given in species ; wherefore the angle BED + 44 Dat. is given: let AE, which is given in magnitude, be given also in position, as also the point E, and the straight line ED will be given * in position : join AD, and because the sum of the squares of AB, BD, that is *, the square of AD is given, therefore the straight * 47. 1. line AD is given in magnitude; and it is also given in position, because from the given point A it is drawn to the straight line ED given in position: therefore the point D, in which the two straight lines AD, ED, given in position, cut one another, is given *: * 28 Dat. and the straight line DB, which is at right angles to AB, is given * in position, and AB is given in posi- * 33 Dat. tion, therefore * the point B is given: and the points * 28 Dat. A, D, are given, wherefore * the straight lines AB, * 29 Dat. BD are given : and the ratio of AB to BC is given, and therefore * BC is given. The Composition. * 32 Dat, 34 Dat. . *2 Dat. AA E BN M. C E given straight line which is to be taken from BC, and let the ratio which the remainder is required to have to BD be the given ratio of HG to GL, and place GL. at right angles to FH, and join LF, LH: next, as HG is to GF, so make HK to AE; produce AE to N, so that AN be the straight line to the square of which the sum of the squares of AB, BD, is required to be equal; and make the angle NED equal to the angle GFL; and from the centre A, at the distance AŇ, describe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN and DM making the angle BDM equal to the angle GLH. Lastly, produce BM to C, so that MC be equal to HK; then is AB the first, BC the second, and BD the third of the straight lines that were to be found. For the triangles EBD, FGL, as also DBM, LGH, being equiangular, as EB to BD, so is FG to GL; and as DB to BM, so is LG to GH; therefore, ex æquali, as EB to BM, so is (FG to GH, and so is) AÈ to HK or MC; wherefore *, AB is to BC, as AE to HK, that is, as FG to GH, that is, in the given ratio: and from the straight line BC taking MC, which is equal to the given straight line HK, the remainder BM has to BD the given ratio of HG to GL: and the sum of the squares of AB, BD, is equal* to the square of AD or AN, which is the given space. * 12. 5. * 47. 1. Q. E. D. I believe it would be in vain to try to deduce the preceding construction from an algebraical solution of the problem. END OF THE NOTES TO THE DATA. PLANE TRIGONOMETRY. LEMMA I. Fig. 1. LET ABC be a rectilineal angle: if about the point B as a centre, and with any distance BA, a circle be described, meeting BA, BC, the straight lines including the angle ABC in A, C; the angle ABC will be to four right angles, as the arch AC to the whole circumference. Produce AB till it meet the circle again in F, and through B draw DE perpendicular to AB, meeting the circle in D, E. By 33. 6. Elem. the angle ABC is to a right angle ABD, as the arch AC to the arch AD; and quadrupling the consequents, the angle ABC will be to four right angles, as the arch AC to four times the arch AD, or to the whole circumference. LEMMA II. Fig. 2. Let ABC be a plane rectilineal angle as before : about B as a centre with any two distances BD, BA, let two circles be described meeting BA, BC, in D, E, A, C; the arch AC will be to the whole circumference of which it is an arch, as the arch DE is to the whole circumference of which it is an arch. By Lemma 1. the arch AC is to the whole circumference of which it is an arch, as the angle ABC is to four right angles; and by the same Lemma 1. the arch DE is to the whole circumference of which it is an arch, as the angle ABC is to four right angles; therefore the arch AC is to the whole circumference of which it is an arch, as the arch DE to the whole circumference of which it is an arch. DEFINITIONS. Fig. 3. I. Let ABC be a plane rectilineal angle ; if about B as a centre, with BA any distance, a circle ACF be de |