E B A Let AB be one of the given G lines ; draw BG to make any angle with AB, and let B D be equal to the other given line. Take AB : BE in the given ratio of the remaining parts, and BF : BE in 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, (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. F 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. M E (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. 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. 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; B H IG but AL : AM is equal to the given ratio, .. also AL : BN is equal to the given ratio. And the same may be proved of any other lines drawn in the same manner. SECT. II. C B F (1.) If a straight line be drawn to touch a circle, IF F and be parallel to a chord; the point of contact will be the middle point of the arc cut off by that chord. Let CD be drawn touching the circle ABE in the point E, and parallel to the chord AB; E is the middle point of the arc AEB. Join AE, EB. The angle BAE is equal to the alternate angle CEA, and therefore to the angle EBA in the alternate segment, whence AE= and (Eucl. iii. 28.) the arc AE is equal to the arc EB. Cor. 1. Parallel lines placed in a circle cut off equal parts of the circumference. If FG be parallel to AB; the arc EF=EG, whence AG=BF. Cor. 2. The two straight lines in a circle, which join the extremities of two parallel chords are equal to each other. For if AB, FG be parallel, the arcs AG, BF are equal, therefore (Eucl. iii. 29.) the straight lines AG, BF are also equal. EB, E B P (2.) If from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the centre, the parts of the lines which are intercepted within the circle are equal. From the point P without the circle ABC let two lines PB, PD be drawn making equal angles with PO, the line joining P and the centre; AB shall be equal to CD. Let fall the perpendiculars OE, OF; then since the angle at E is equal to the angle at F, and EP0=FPO, and the side PO, opposite to one of the equal angles in each is common, :: OE=OF, and consequently (Eucl. iii. 14.) AB=CD. F B (3.) Of all straight lines which can be drawn from two given points to meet on the convex circumference of a given circle ; the sum of those two will be the least, which make equal angles with the tangent at the point of concourse. Let A and B be two given points, CE a tangent to the circle at C, where the lines AC, BC make equal angles with it; and let lines AD, BD be drawn from A and B to any other point D on the convex circumference; AC and CB together are less than AD, DB together. D E |