fore, because the straight line DC is drawn to the given point D, in the straight line BD, given in position, in the given c 32. dat. angle BDC, DC is given in position: and the circumference d 28. dat. ABC is given in position; therefored the point C is given. If from a given point a straight line be drawn touching a circle given in position ; the straight line is given in position and magnitude. Let the straight line AB be drawn from the given point A, touching the circle BC, given in position: AB is given in position and magnitude. Take D the centre of the circle, and join DA, DB: Because each of the points D, A is given, B a 29. dat. the straight line A Disgivena in position and magnitude: And DBA is a right b c Cor.5.4. angle, wherefore DA is a diameter of the circle DBA, described about the D A triangle DBA: and that circle is thered 6. def. fore given d in position: And the circle BC is given in position, therefore the e 28. dat. point B is given The point A is also given: therefore the straight line AB is given a in position and magnitude. b 18. 3. If a straight line be drawn from a given point without a circle, given in position ; the rectangle contained by the segments betwixt the point and the circumference of the circle is given. Let the straight line ABC be drawn from the given point A, without the circle BCD, given in position, cutting it in B, C; the rect D angle BA, AC is given. From the point A draw a AD touchb 94. dat. ing the circle; therefore AD is given C Cl в А in position and magnitude : And be cause AD is given, the square of AD c 56. dat. is given “, which is equal d to the rect angle BA, AC: Therefore the rectangle BC, AC is given. a 17. 3. d 36. 3. If a straight line be drawn through a given point within a circle, given in position, the rectangle contained by the segments betwixt the point and the circumference of the circle is given. Let the straight line BAC be drawn through the given point A within the circle BCE, given in position; the rectangle BA, AC is given. Take D the centre of the circle, join AD, and produce it to the points E, F: E Because the points A, D are given, the straight line AD is given a in position ; D and the circle BEC is given in position; B с therefore the points E, F are given b; A and the point A is given, therefore EA, T AF are each of them given a ; and the rectangle EA, AF is therefore given ; and it is equal o to the c 35. 3. rectangle BA, AC, which consequently is given. a 29. de b 28 d If a straight line be drawn within a circle given in magnitude, cutting off a segment containing a given angle ; if the angle in the segment le bisected by a straight line produced till it meets the circumference, the straight lines which contain the given angle shall both of them together have a given ratio to the straight line which bisects the angle : And the rectangle con. tained by both these lines together which contain the given angle, and the part of the bisecting line cut off below the base of the segment, 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 BAC be bisected by the straight line AD; BA together with AC has a given ratio to AD; and the rectangle contained by BA and AC together, and the straight line ED cut off from AD below BC, the base of the segment is given. Join BD; and because BC is drawn within the circle ABC e 21. 3. given in magnitude, cutting off the segment BAC, containing a 91. dat. the given angle BAC; BC is given in magnitude: By the b ). dat. same reason, BD is given; therefore b the ratio of BC to BD is given: And because the angle BAC is bisected by AD, as c 3. 6. BĂ to AC, so is © BE to EC; and, by permutation, as AB d 12. 5. to BE, so is AC to CE; wherefore, as BA and AC together to BC, so is AC to CE: And because the angle BAE is equal A B C D Also the rectangle contained by BA and AC together, and Because the triangle BDE is equiangular to the triangle ACE, as BD to DE, so is AC to CE; and as AC to CE, SO is BA and AC to BC; therefore as BA and AC to BC, so is BD to DE; wherefore, the rectangle contained by BA and AC together, and DE, is equal to the rectangle CB, BD; But CB, BD is given : therefore the rectangle contained by BA and AC together, and DE is given. Otherwise. a Produce CA, and make AF equal to AB, and join BF; 55. & and because the angle BAC is the double a of each of the an( 32. 1. gles BFA, BAD, the angle BFA is equal to BAD; and the angle BCA is equal to BDA, therefore the triangle FCB is equiangular to ABD: As therefore FC to CB, so is AD to DB; and, by permutation, as FC, that is, BA and AC together, to AD, so is CB to BD: And the ratio of CB to BD is given; therefore the ratio of BA and AC to AD is given. And because the angle BFC is equal to the angle DAC, that is, to the angle DBC, and the angle ACB equal to the angle ADB; the triangle FCB is equiangular to BDE; as therefore FC to CB, so is BD to DE; therefore the rectangle contained by FC, that is, BA and AC together, and DE is equal to the rectangle CB, BD, which is given, and therefore the rectangle contained by BA, AC together, and DE is given. If 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 cir. cumference, shall be given. a 9). dat. 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 seg D ment BAC, containing a given angle; BC is therefore given a in A F C E G 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 A B 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 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: consequently, 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 CB 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. If from a 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 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: a Cor.5.4. Because GEA is a right angle, the straight line AG is a the diameter of the circle ABC; and BC is also a diameter of it; |