DC is drawn to the given point D in the straight line BD given in c. 32. Dat. position in the given angle BDC, DC is given in position, and the d. 28. Dat.circumference ABC is given in position, therefore the point C is given. 91. PRO P. XCIV. 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 center of the cirele, and join DA, DB. because each of the points D, A is given, the straight B a. 29. Dat. line AD is given in position and magni b. 18. 3. tude. and DBA is a right b angle, wheree.Cor. 5.4. fore DA is a diameter of the circle DBA described about the triangle DBA; and А d. 6. Def. that circle is therefore given d in position. and the circle BC is given in position, theree. 28. Dat. fore the point B is given the point A is also given; therefore the straight line AB is given in position and magnitude. D 92. PRO P. XCV. 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 with out the circle BCD given in position, cutting it in B, C; the rectangle BA, AC is given. From the point A draw * AD touching C BA b. 94. Dat. the circle; therefore AD is given b in pofi tion and magnitude. and because AD is €. 56. Dat. given, the square of AD is given which is d. 36. 3. equal d to the rectangle BA, AC. therefore the rectangle BA, AC is given. PROP. PRO P. XCVI. 93 IF a a straight line be drawn thro' 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 thro' the given point A within the circle BCE given in position; the rectangle BA, AC is given. Take D the center of the circle, join AD and produce it to the points E, F, because E the points A, D are given, the straight line AD is given in position; and the circle BEC 2. 29. Dat. is given in position ; therefore the points E, B C F are given b. and the point A is given, b. 28. Dat. therefore EA, AF are each of them given ; and the rectangle EA, AF is therefore given; and it is equal to c. 35. 3. the rectangle BA, AC which consequently is given. If PROP. XCVII. 94. a straight line be drawn within a circle given in mag nitude cutting off a segment containing a given angle; if the angle in the segment be 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 contained 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 F straight line AD; BA together with AC has a given ratio to AD; and the rect. angle contained by BA and AC together, and the straight line ED cut off from AD below BC the base of the segment, is gi BK JE ven. c. 3. 6. C Join BD; and because BC is drawn within the circle ABC given in magnitude cutting off the segment BAC containing the given a. 91. Dat. angle BAC; BC is given ' in magnitude, by the same reason BD b. 1. Dat. is given ; therefore b the ratio of BC to BD is given. and because the angle BAC is bifected by AD, as BA to AC, fo is BE to EC; d. 12. s. and, by permutation, as AB to BE, fo is AC to CE ; wherefored as BA and AC together to BC, so is AC to CE. and because the angle BAE is equal to EAC, and the F c. 21. 3. angle ACE to ADB; the triangle ACE is equiangular to the triangle ADB; E B D Also the rectangle contained by BA and AC together, and DE is given. Becaufe the triangle BDE is equiangular to the triangle ACE, as AD to DE, fo 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. Produce CA and make AF equal to AB, and join BF. and be2.5. and 32.cause the angle BAC is double * of each of the angles BFA, BAD, the angle BFC is equal to BAD; and the angle BCA is equal to BDA, therefore the triangle FCB is equiangular to ADB. as there. fore FC to CB, fo is AD to DB, and, by permutation, as FC, that is BA and AC together to AD, fo 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, fo 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. PROP. 1. F a straight line be drawn within a circle given in mag nitude 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, thall be given. A 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 thro' B draw BG parallel to DE meeting AC produced in G. and because BC cuts off from the cir cle ABC given in magnitude the fegment BAC containing a given angle, BC is therefore given in magnitude. by the fame rea a. 91. DR. son BD is given, because the angle BAD is equal to the given angle E AF; therefore the ratio of BC to BD is given. and because the angle CAE is equal to E AF, of which G CAE is equal to the alternate angle AGB, and EAF to the ind terior and opposite angle ABG; therefore the angle AGB is equal to A BG, 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 equi Ff angular F A angular to BDA. therefore as GC to CB, fo is AD to DB, and, by And because the angle GBC is equal B angle BCG equal to BDE; the tri C с E angle BCG is equiangular to BDE. therefore as GC to CB, so is BD to G DE, and consequently the rectangle GC, DE is equal to the rect. angle 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 froin the point in which this ineets 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 4. Cor. 5.4.GEA is a right angle, the straight line AG is the diameter of the circle ABC; and BC is also a diameter of it; therefore the point H where they meet is the center of the circle, and consequently H is given. and the point D is given, wherefore DH is given in magni. tude. |