remainder is required to have to BD, be the given ratio of HG to LG, 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 AN 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) AE to HK or MC; wherefore -(12. 5.) 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 (47. 1.) to the square of ĄD or AN, which is the given space. Q. E. D. I believe it would be in vain to try to deduce the preceding construction from an algebraical solution of the problem. FINIS. 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, aş 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 ciroles 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 described, meeting BA, BC in A, C; the arch AC is called the measure of the angle ABC. The circumference of a circle is supposed to be divided into 360 equal parts called degrees; and each degree into 60 equal |