PROP IX. THEOR. If two straight lines in a given ratio, and cotaining a given angle, terminate in two diverging lines which are given in position, the locus of their vertex will be likewise a straight line given in position. Let the straight lines AB, AC, in a given ratio, and containing a given angle, be limited by the given diverging lines DE, DF; then will their vertex A lie in a given direction. M Join DA, and produce BA to meet DF in G. The triangle DBG is given in species; for the angles at D and B are given, and, consequently, the angle at G. Again, the triangle ACG is given in species, since all its angles are given. Hence the ratio of AC to AG is given; but the ratio of AB to AC is given, and consequently that of AB to AG and that of BG to AG. H N IC GF Hence, also, the ratio of BG to DG is given, and therefore the ratio of AG to DG; and the angle at G being given, the triangle DAG is (VI. 15. El.) consequently given in species. Wherefore the angle GDA is given, and hence the straight line DA is given in position. COMPOSITION. In DE take any point H, and draw HI and HL, making with DE and DF angles equal to the respective inclinations of the bounded lines, produce IH to M, so that MH shall have to HL the given ratio, find IN a third proportional to IM, IH, and join DNA; this straight line is the locus required. Because IM: IH :: IH: IN, therefore (V. 11. and 7. El.) MH: IM :: NH: IH; but (VI. 2. El.) AB : AG :: HN: IH, and the triangles ACG and HLI being evidently similar, AG: AC:: IH: HL; therefore (V. 16. El.) AB: AC:: MH: HL, that is, in the given ratio. PROP. X. THEOR. Three diverging lines being given in position, if a straight line cut them at given angles, and such that the rectangle of its first segment, by a given line, shall be equal to both the rectangles of its second and third segments by given lines; the locus of its point of origin will be a straight line given in position. Let ABCD cut the diverging lines EF, EG and EH at given angles, and let AB.KL=AC.ML+AD.NM; then will the locus of the point A be a straight line given in position. Because AC.ML ANALYSIS. AB.ML+ BC.ML, and AD.NM = AB.NM+ BD.NM, therefore AB. KL = AB. ML + BC.ML+AB.NM+BD.NM,and consequently AB.KL= AB(ML+NM)+BC.ML+ B G BD. NM, and AB. KN = BC.ML+BD.NM. Make BC: BD :: NM: MO, and BC.MO=BD.NM; whence AB.KNBC(ML+MO) = BC.OL, and AB: BC:: OL: KN. The ratio of AB to BC is, therefore, given; but the triangle BCE being E given in species, the ratio of BE to BC is given, and consequently the ratio of AB to BE is given; and since the contained angle ABE is given, the triangle BEA is likewise given in species; and thence the point A, and the straight line EA, are given in position. K COMPOSITION. D H ON M Having assumed in EH any point H, draw HGF in the given inclination, make FG : GH :: NM: MO, and produce HF till KN : OL :: FG : IF; EI is the straight line required. For BC: AB :: FG: IF :: KN: OL, and AB.KN =BC.OL; but BC: CD:: FG: GH:: NM: MO, and BC. MO CD. NM. Wherefore AB. KN= BC. OL= BC.ML+CD.NM, and AB.KM= AB.NM + BC.ML + CD. NM BC. ML + AD. NM, and hence AB. KL = = MLAD. AB. ML+BC. ML + AD. NM AC. ML + AD. NM. BOOK 111. PROP. XI. THEOR. Four diverging lines being given in position, if a straight line cut them at given angles, and such that the rectangles of its first and second segments by given lines shall be equal to both the rectangles of its third four segments by given lines; the locus of its point of origin will be a straight line given in position. Let ABCDE cut the diverging lines FG, FH, FI, and FK at given angles, and let AB.MN+AC.NO = AD.OP +AE.PQ; then will the locus of the point A be a straight line given in position. ANALYSIS. Because AB.MN+AC.NO AD.OP+AE.PQ, it follows, by decomposition, that AB.MO+BC.NO=AB.OQ +BD.OP+BE.PQ, and con sequently AB.MQ + BC.NO =BD.OP+BE.PQ. Make BD: BC :: NO: OR, and M AB to BF is given, and the contained angle ABF being given, the triangle BFA is likewise given in species; and hence the straight line FA is given in position. COMPOSITION. Having assumed in FK any point K, draw KIHG at the given inclination, make GI: GH:: NO: OR, and GI: GK :: PQ PS, and produce KG till MQ: SR:: GI: GL; FL is the straight line required. For BD: BC: GI: GH :: NO: OR, and BD.OR = BC.NO; but BD: BE :: GI: GK :: PQ : PS, and BD.PS BE.PQ; again, MQ: SR :: GI : GL :: BD : AB, and AB.MQ=BD.SR. Whence AB.MQ+BC.NO=BD.SR + BD.OR = BD.SO = BD.PS + BD.OP = BE.PQ + BD.OP. PROP. XII. THEOR. If a straight line given in position, be cut at given angles by two straight lines, which intercept, from two given points in it, segments that have a given ratio, the locus of the point of concourse is a straight line given in position. Let AB and AC be drawn, such that the angles ABF, and ACF, with the ratio of DB to EC, are given; the locus of A, the point of concourse, is a straight line given in position. |