For because AC: CB:: AD: DB and EA BG: AC: CB, (by sim. tri. ACE, BCG) : .. (Eucl. v. 11.) EA : BG :: DA : DB, (Eucl. vi. 6.) the triangles EAD, GDB are equiangular, and ED: DG:: AE: BG: CE: CG. (29.) From two given points, to draw two straight lines which shall contain a given angle, and meet two lines given in position, so that the parts intercepted between those points and the lines may have a given ratio. Let AB, CD be the lines given in position, and E, F the given points. From E draw EA perpendicular to AB, and make the angle AGF equal to the given angle. In GF produced take FH such, that the ratio of EA -: FH may be the same as the given ratio. Draw HD perpendicular to GH meeting CD in D. Draw DFI E K H and BEI to include the given angle. These are the lines required. For, since the angles FGE, FIE are equal, as also FKG, EKI, .. GFK, IEK or their vertically opposite. angles DFH, AEB are equal, and the angles at H and A are right angles, .. the triangles FDH, AEB are equiangular, and EB FD: EA: FH, i. e. in the given ratio. (30.) The length of one of two lines which contain a given angle being given; to draw from a given point without them a straight line which shall cut the given line produced, so that the part produced may be in a given ratio to the part cut off from the indefinite line. in the given ratio. Divide DF so that FE: DG :: FG AB. Join AG; and draw DH parallel to AG, and it will be the line cutting BC in H, and BA produced in I, as was required. Join AF; and draw BK parallel to AG cutting AF in L; and draw LM parallel to KE cutting AE in M and AG in N. Then FE LM :: GF: (NL=) AB : and FE DG :: FG AB by construction; : : .. LM=DG=IA; if therefore ILO be drawn, IL must be equal and parallel to AM, and IO to AE (Eucl. i. 33.). In the same manner it is evident that HB= IL= AM ; and by similar triangles AFE, ALM, FE EA: LM: MA :: IA: HB .. IA: HB in the given ratio. (31.) From two given straight lines to cut off two parts, which may have a given ratio; so that the ratio of the remaining parts may also be equal to the ratio of two other given lines. D Let AB be one of the given lines; draw BG to make any angle E with AB, and let BD be equal to the other given line. Take AB : BE in the given ratio of the remaining parts, and BF: BE in A the given ratio of the parts to be cut off. Join AE, FE; and draw DH and BC parallel to EF, and HC parallel to DB meeting BC in C, and AB in I. Then (Eucl. vi. 2.) AI : IH :: AB : BE in the given ratio of the remainders; and the triangles BCI, BFE having the angle CBI= the alternate angle BFE, and CIB=FBE, are equiangular, .. BI : IC :: BF : BE, in the ratio of the parts to be cut off; and AB, HC (=DB) are the given lines. (32.) Three lines being given in position; to determine a point in one of them, from which if two lines be drawn at given angles to the other two, the two lines so drawn may together be equal to a given line. Let AB, AC, BC be the three lines given in position, take AD = the given line, and making with AB an angle equal to one of the given angles. Through D draw Dba parallel to AB, and meeting AC and BC in a and b. Draw AE to meet CB in E making the angle AEC= the given angle to be made by the line to be drawn, with BC. In AE take Ad=AD, and join ad cutting BC in F. Draw FG parallel to EA meeting AC in G, which is the point required. For through G draw IGH parallel to DA, then the triangles a GI, a AD are similar, and aA; AD=Ad: aG: GI; but a A Ad: aG: GF, and .. GI=GF, .. GH+GF=GH+GI = AD=the given line; and the angle GHB=DAB, and GFC= AEC,.. GHB, GFC are equal to the given angles. (33.) If from a given point two straight lines be drawn including a given angle, and having a given ratio,, and one of them be always terminated by a straight line, given in position; to determine the locus of the extremity of the other. Let A be the given point, and BC the line given in position. From A draw any line AD, and make the angle DAE equal to the given angle, and take AE such that AD: AE may be in the given ratio; and through E draw EF making the angle AEF=ADB; EF is the locus required. Draw any other line AB, and make the angle BAF =DAE. Then the angle BAD=FAE and ADB= AEF, .. the triangles ABD, AEF are equiangular, whence AB AF :: AD: AE, in the given ratio. The same may be proved of any other lines drawn from A and containing an angle equal to the given angle, and one of them terminated in BC. (34.) If from two given points, straight lines be drawn, containing a given angle, and from each of them segments be cut off, having a given ratio; and the extremities of the segments of the lines drawn from one of the points be in a straight line given in position; to determine the locus of the extremities of the segments of lines drawn from the other. C B M L D E 4 Let A and B be the given points, and CD the line given in position. From A to CD draw any line AE. Make the angle EAF the given angle, and AE : AF in the given ratio, and let FG be the locus of the points F (i. 33.). Draw BH equal and parallel to AF, and through H draw HI parallel to GF. It is the locus required. H /G Draw any lines AK, BK containing the angle at K =the given angle. Make the angle LAM=the given angle; AL: AM in the given ratio, and M is in the line GF. And since AF is parallel to BH, and FM to HN, and BK to AM (since the angles BKA, LAM are equal) and AF = BH, .. the triangles BHN, AFM are similar and equal, .'. AM=BN; |