B adjacent to BAC be bisected by the straight line DAE, meeting the circumferenee again in D, and BC the base of the segment produced in E; the excess of BA, AC, has a given ratio to AD; and the rectangle which is contained by the same excess and the straight line ED is given. Join BD, and through B, draw BG parallel to DE meeting AC produced in G: and because BC cuts off from the circle ABC given in , magnitude, the segment BAC containing a given an* 91 Dat. gle, BC is therefore given * in magnitude : by the same And because the angle GBC is equal to the alternate angle DEB, and the angle BCG equal to BDE; the triangle BCG is equiangular to BDE: therefore as GC to CB, so is BD to DE; and consequently the rectangle GC, DE is equal to the rectangle CB, BD which is given, because its sides CB, BD are given : therefore the rectangle contained by the excess of BA, AC, and the straight line DE is given. If from a given point in the diameter of a circle given in position, or in the diameter produced, a straight line be draron to any point in the circumference, and from that point a straight line be drawn at right angles to the first, and from the point in which this meets the circumference again, a straight line be drawn parallel to the first; the point in which this parallel meets the diameter is given ; and the rectangle contained by the two parallels is given. In BC the diameter of the circle ABC given in position, or in BC produced, let the given point D be taken, and from D let a straight line DA be drawn to any point A in the circumference, and let AE be drawn at right angles to DA, and from the point E where it meets the circumference again let EF be drawn parallel to DA meeting BC in F; the point F is given, as also the rectangle AD, EF. Produce EF to the circumference in G, and join AG: because GEA is a right angle, the straight line AG is * the diameter of the circle ABC; and BC is * Cor. 5. 4. also a diameter of it; therefore the point H, where they meet, is the centre of the circle, and consequently H is given :-and the point D is given, wherefore DH is given in magnitude. And because AD is parallel to FG, and GH equal to HA; DH is equal * to HF, * 4.6. and AD equal to GF: and DH is given, therefore HF is given in magnitude; and it is also given in po- . A С Е E G sition, and the point H is given, therefore * the point * 30 Dat. F is given. And because the straight line EFG is drawn from a given point F without or within the circle ABC given in position, therefore * the rectangle EF, FG is given: Dat. and GF is cqual to AD, wherefore the rectangle AD, EF is given. F F * 95 or 96 If from a given point in a straight line given in position, a straight line be drawn to any point in the circumference of a circle given in position ; and from this point a straight line be drawn, making with the first an angle equal to the difference of a right angle, and the angle contained by the straight line given in position, and the straight line which joins the given point and the centre of the circle ; and from the point in which the second line meets the circumference again, a third straight line be drawn, making with the second an angle equal to that which the first makes with the second : the point in which this third line meets the straight line given in position is given ; as also the rectangle contained by the first straight line, and the segment of the third betwixt the circumference and the straight line given in position, is given. D WB M Let the straight line CD be drawn from the given point C, in the straight line AB given in position, to the circunference of the circle DEF given in position, of which Ğ is the centre; join CG and from the point D let DF be drawn making the angle CDF equal to the difference of a right angle, and the angle BCG, and from the point Flet FE be drawn, making the angle DFE, equal to CDF, meeting AB in H: The point H is given ; as also the rectangle CD, FH. Let CD, FH, meet one another in the point K, from which draw KL perpendicular to DF; and let DC meet the circumference again in M, and let FH meet the same in , and join MG, GF, GH. Because the angles MDF, DFE, are equal to one another, the circumferences MF, DE, are equal*; and * 26. 3. adding or taking away the common part ME, the circumference DM is equal to EF; therefore the straight line DM is equal to the straight line EF, and the angle GMD to the angle * GFE; and the angles GMC, * 8. 1. GFH are equal to one another, because they are either the same with the angles GMD, GFE, or adjacent to them : and because the angles KDL, LKD, are together equal * to a right angle, that is, by thé * 32. 1. hypothesis, to the angles KDL, GCB; the angle GCB or GCH is equal to the angle (LKD, that is, to the angle) LKF or GHK: therefore the points C, K, H, G, are in the circumference of a circle: and the angle GCK is therefore equal to the angle GHF: and the angle GMC is equal to GFH, and the straight line GM to GF; therefore * CG is equal to GH, and CM * 26. 1. to HF: and because CG is equal to GH, the angle GCH is equal to GHC; but the angle GCH is given: therefore ĜHC is given, and consequently the angle CGH is given ; and CG is given in position, and the point G; therefore * GH is given in position; and * 32 Dat. CB is also given in position, wherefore the point H is given. And because HF is equal to CM, the rectangle DC, FH, is equal to DC, CM: but DC, CMR is given *, because the point C is given, therefore the * 95 or 96 Dat. rectangle DC, FH is given. END OF THE DATA. |