DC is drawn to the given point D in the straight line DD given C. 32. Dat. in position in the given angle BDC, DC is given in position. and d. 28. Dat. the circumference ABC is given in position, therefore 4 the point C is given. PRO P. XCIV. ing 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 circle, and join DA, DE. because each of the points D, A is given, the straight 2. 29. Dat. line AD is given in position and magni tade. and DBA is a right bangle, wherec. Cor. 5.4. fore DA is a diameter of the circle DBA described about the triangle DBA; and D A d. 6. Def. that circle is therefore given a iu 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. b.rt. 3. 92. PROP. XCV. out a circle given in pofition ; the rectangle contained by the fogments 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 ECD given in politior, cutting it in B, C; the rectangle BA, AC is given. From the point A draw' AD touching C b. 94. Dat. the circle; therefore AD is given b in po sition and magnitude. and because AD is C. 56. Dat. given, the square of AD is given which is equal d to the rectangle BA, AC. therefore the rectangle EA, AC is given. a. 17. 3. BA d. 36. 3: IF PROP. XCVI. S3 . F a straight line be drawn thro’a given point witliin 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. Tike D the center of the circle, join AD and produce it to the points E, F. because the points A, D are given, the straight line AD is given in position; and the circle BEC a. 29. Dat. is given in pofition; therefore the points E, B CA F are given b. and the point A is given, F 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 confequently is given. B b. 28. Dat. PRO P. XCVII. 94. IF F a straight line be drawn within a circle given in magnitude cutting off a fegment containing a given angle ; if the angle in the segment be bisected by a straight line produced till it meets the circumference, the firaight lines which contain the given angle shall both of them together have a given ratio to the straight line which bisecis 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 DC be drawn within the circle ABC given in magnitude cutting off a segment containing the given anglc BAC. and let the angle BAC be birected by F the straight line AD; BA together with AC has a given ratio to AD, and the rectangle contained by DA and AC together, and the straight line ED cut off from AD below BC the base of the fogment, is givea. ci 3. 6. d. 12. 5. Join BD; and because BC is drawn within the circle ABC given iš magnitude cutting off the fegment BAC containing the given 1. 91. Dat. angle BAC; BC is given in magnitude. by the fame 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, so is BE to EC; and, by permutation, as AB to BE, so is AC to CE; wherefore as BA and AC together to BC, fo is AC to CE. and because the angle BAL is equal to EAC, and the F A 6. 21. 3. angle ACE to o ADB ; the triangle ACE E B D Also the rectangle contained by BA and AC together, and DE is given. Because the triangle BDE is equiangular to the triangle ACE, as BD to DE, fo is AC to CE; and as AC to CE, fo 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. Otherwisc. 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 therefore 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 equiangalar 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. 1. F a fraight line be drawn within a circle giren in magnitude cutting oila legment containing a givet angle; it the angle adjacent to the angle in the segment be bileted by a ftraight line produced till it mect the circumference again and the base of the sevinent; the excess of the straight lines which contain the given angle ihall have a given ratio to the segment of the bifesting line which is within the circle ; and the reciangle contained by the same exceis and the fegment of the bifecting line betwixt the bare produced and the point where it again meets the circumference, thall be given. Let the straight line BC be drawn within the circle ADC giren in magnitude cutting off a fegment containing the given angle BAC, and let the angle CAF adjacent to BAC be bitected by the straight line DAE meeting the circumference again in D, and BC the baie of the segment produced in E; the excess of PA, AC has a given ratio to AD, and the rectangle which is contained by the fanie excess and the straight line ED, is given. Join BD, and thro' B draw BG parallel to De meet'ng AC :0duced in G. and because BC cuts off from the circle ABC given in magnitude the fegment BAC contain D ing a given angle, BC is therefore given ' in magnitude. by the same rea a. 91. IM fon BD is given, hecaule the angle BAD is equal to the given angle EAF; therefore the ratio of BC to BD is gi-B C D ren. and because the angle CAE is eequal to EAF, of which CAE is equal to the alternate angle AGB, and EAF to the interior and oppofire angle ABG; therefore the angle AGB is equal to ABG, and the straight line AB equal to AG; so that GC is the exceis 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 qua rilateral BCAD in the circle; therefore the triangle BGC is equiangular to DDA Ff thercfore as CC to CB, fo is AD to DB, and, by permutation, as And because the angle GBC is e- 95. PRO P. XCIX. . I F from a given point in the diameter of a circle given in position, or in the diameter produced, a ftraight line be drawn to any point in the circumference, and from that point a itraight line be drawn at right angles to the first, and from the point in which this mects 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 EC in F; the point F is given, as also the rectangle AD, EF. Produce EF to the circumference in G, and join AG. because * 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 giveo |