DE is equal to the rectangle CB, BD, which is given, and therefore the rectangle contained by BA, AC together, and DE, is given. Ir a straight line be drawn within a circle given in magnitude, cutting off a segment containing a given angle: If the angle adjacent to the angle in the segment be bisected by a straight line produced till it meet the circumference again, and the base of the segment; the excess of the straight lines which contain the given angle shall have a given ratio to the segment of the bisecting line which is within the circle; and the rectangle contained by the same excess, and the segment of the bisecting line betwixt the base produced and the point where it again meets the circumference shall be given. Let the straight line BC be drawn within the circle ABC given in magnitude, cutting off a segment containing the given angle BAC, and let the angle CAF adjacent to BAC be bisected by the straight line DAE, meeting the circumference 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 angle, BC is therefore 91 Dat. given a in magnitude: By the same reason BD is given, because the angle BAD is equal to the given angle EAF; therefore the B ratio of BC to BD is given: And because the angle CAE is equal to EAF, of which CAE is equal to the alternate angle AGB, and EAF to the interior and opposite angle ABG; therefore the angle AGB is equal to ABG, and the straight line AB equal to AG; so that GC is the excess of BA, AC: And because the angle BGC is equal to GAE, that is, to EAF, or the angle BAD; and that the angle BCG is equal to the opposite interior angle BDA of the quadrilateral BCAD in the circle; therefore the triangle BGC is equiangular to BDA. Therefore as GC to CB, so is AD to DB; and, by permutation, as GC, which is the excess of BA, AC, to AD, so is BC to BD: And the ratio of CB to BD is given; therefore the ratio of the excess of BA, AC, to AD is given. 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. PROP. XCIX. Ir from the given point in the diameter of a circle given in position, or in the diameter produced, a straight line be drawn 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 ią 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. 95. Produce EF to the circumference in G, and join AG: Because GEA is a right angle, the straight line AG is the Cor. 5. 4 diameter of the circle ABC; and BC is 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 Dis given, wherefore DH is given in magnitude. And because AD is parallel to FG, and GH equal to HA; DH is 4. 6. equal to HF, and AD equal to GF: And DH is given, therefore HF is given in magnitude; and it is also given 30 Dat. in position, and the point H is given, therefore the point 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, 95 or 96 therefored the rectangle EF, FG is given: And GF is equal to AD, wherefore the rectangle AD, EF is given. 1 Dat. 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.A Let the straight line CD be drawn from the given point C in the straight line AB given in position, to the circumference of the circle DEF given in position, of which G 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 F let FE be drawn, making the angle DFE equal to CDF, meeting AB C D A F B M ⚫ 26.3. K b 8. 1. K 32. 1. Because the angles MDF,DFE, are equal to one another, the circumferences MF, DE, are equal; and 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 angleb GFE; and the angles GMC, GFH are equal to one another, because E 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 the hypothesis, to the angles KDL, GCB; the angle GCB or GCH is equal to the angle (LKD, that is, to the angle) LKF or GKH: 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; therefored CG is equal to GH, and CM to HF: And be- 26. 1. cause CG is equal to GH, the angle GCH is equal to GHC; but the angle GCH is given: Therefore GHC is given, and consequently the angle CGH is given; and CG is given in position, and the point G; therefore e GH is given in posi- 32 Dat. tion; and CB is also given in position, wherefore the point H is given. M A C HB e And because HF is equal to CM, the rectangle DC, FH, is equal to DC, CM: But DC, CM is given, because the f 95 or 96 point C is given, therefore the rectangle DC, FH is given. Dat. END OF THE DATA. |