a Join BD; and because BC is drawn within the circle ABC given in magnitude cutting off the fegment BAC containing the given a. 91. Dat. angle BAC; BC is given in magnitude. by the fame reafon BD b. 1. Dat. is given; therefore the ratio of BC to BD is given. and because c. 3. 6. the angle BAC is bifected by AD, as BA to AC, fo is BE to EC; d. 12. 5. and, by permutation, as AB to BE, fo is AC to CE; wherefore d as BA and AC together to BC, fo is AC to CE. and because the angle BAE is equal to EAC, and the F e. 21. 3. angle ACE to ADB; the triangle ACE A is equiangular to the triangle ADB; E B C D I. tation, as BA and AC to AD, fo is BC to BD. and the ratio of BC to BD is given, therefore the ratio of BA together with AC to AD is given. 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, fo 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. Otherwife. Produce CA and make AF equal to AB, and join BF. and bea. 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 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. IF Fa ftraight line be drawn within a circle given in magnitude cutting off a fegment containing a given angle; if the angle adjacent to the angle in the fegment be bifected by a ftraight line produced till it meet the circumference again and the base of the fegment; the excefs of the straight lines which contain the given angle fhall have a given ratio to the fegment of the bifecting line which is within the circle; and the rectangle contained by the fame excefs and the fegment of the bifecting line betwixt the bafe produced and the point where it again meets the circumference, fhall be given. Let the ftraight line BC be drawn within the circle ABC given in magnitude cutting off a fegment containing the given angle BAC, and let the angle CAF adjacent to BAC be bifected by the straight line DAE meeting the circumference again in D, and BC the base of the fegment produced in E; the excess of BA, AC has a given ratio to AD; and the rectangle which is contained by the fame excefs and the ftraight line ED, is given. D P. F a. 91. Dat. Join BD, and thro' B draw BG parallel to DE meeting AC produced in G. and because BC cuts off from the circle ABC given in magnitude the fegment BAC containing a given angle, BC is therefore given in magnitude. by the fame reafon BD is given, because the angle BAD is equal to the given angle EAF; therefore the ratio of BC to BD is gi-B ven. and because the angle CAE is equal to EAF, of which CAE is equal G to the alternate angle AGB, and EAF to the interior and oppofite angle ABG; therefore the angle AGB is equal to ABG, and the ftraight line AB equal to AG; so that GC is the excefs 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 oppofite interior angle BDA of the quadrilateral BCAD in the circle; therefore the triangle BGC is equiangular to BDA. Ff 7 I Let the ftraight line BC be drawn within the circle ABC given Join BD, and thro' B draw BG parallel to DE meeting Arm ing a given angle, BC is therefore gi the alternate angle AGB, and EF T L is aight rcumfe As given. ven point C in the imference of the circle center; join CG, and 450 therefore as GC to CB, fo is AD to DB, and, by permutation, as And because the angle GBC is D F the angle BCG equal to BDE; the tri- B C E G 95. DE, and confequently the rectangle GC, DE is equal to the rectangle CB, BD which is given, because its fides CB, BD are given. therefore the rectangle contained by the excefs of BA, AC and the straight line DE is given. PROP. XCIX. IF from a given point in the diameter of a circle given in pofition, or in the diameter produced, a ftraight line be drawn to any point in the circumference, and from that point a ftraight 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 firft; 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 pofition, or in BC produced, let the given point D be taken, and from D let a ftraight 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 a. Cor. 5.4. GEA is a right angle, the ftraight line AG is the diameter of the circle ABC; and BC is alfo a diameter of it; therefore the point H where they meet is the center of the circle, and confequently H is given. and the point D is given, wherefore DH is given |