of one of the sides of the parallelogram, together with the space which has a given ratio to the square of the other side is to be made equal; and let thïs given ratio be the same wbich the square of the given straight line MG has to the square of GF. By the 88th dat. find two straight lines DB, BC which contain a rectangle equal to the given rectangle MG, GK, and such that the sum of their squares is equal to the given rectangle KG, GL; therefore, by the determination of the problem in that proposition, twice the rectangle MG, GK must not be greater than the rectangle KG, GL. Let it be so, and join the straight lines DB, BC in the angle DBC equal to the given angle FGH; and, as MG to GF, so make DB to BA, and complete the parallelogram AC: AC is equal to the rect BE K L angle FH, GK; and the square of BC together with the square of BD, which, by the construction, has to the square of BA the given ratio which the square of MG has to the square of GF, is equal, by the construction, to the given rectangle KG, GL. Draw A E perpendicular to BC. Because, as DB to BA, so is MG to GF; and as BA to AE, so GF to FH; ex æquali, as DB to A E, so is MG to FH; therefore, as the rectangle DB, BC to AE, BC, so is the rectangle MG, GK to FH, GK; and the rectangle DB, BC is equal to the rectangle MG, GK: therefore the rectangle AE, BC, that is, the parallelogram AC, is equal to the rectangle FH, GK. If a straight line drawn within a circle given in magnitude cuts off a segment which contains a given angle ; the straight line is given in magnitude. In the circle ABC given in magnitude, let the straight line AC be drawn, cutting off the segment AEC, which contains the given angle AEC; the straight line AC is given in magnitude. Take D the centre of the circle “, join AD, and produce it a ). 3. 90. If a straight line given in magnitude be drawn within a circle given in magnitude, it shall cut off a segment containing a given angle. Let the straight line AC, given in magnitude, be drawn within the circle ABC, given in magnitude; it shall cut off a segment containing a given angle. Take D the centre of the circle, join AD, and produce it to E, and join EC: B And because each of the straight lines E EA and AC is given, their ratio is D given a ; and the angle ACE is a right a 1. dat. angle, therefore the triangle ACE is A с given in species, and consequently the b 46. dat. angle AEC is given. PROP. XCIII. If from any point in the circumference of a circle given in position two straight lines be drawn, meeting the circumference, and containing a given angle ; if the point in which one of them meets the circumference again be given, the point in which the other meets it is also given. From any point A in the circumference of a circle ABC given in position, let AB, AC be drawn to the circumference, making the given angle BAC; if the point A B be given, the point C is also given. Take D the centre of the circle, and join BD, DC; and because each of the D points B, D is given, BD is given a in B a 29. dat. position; and because the angle BAC is given, the angle BDC is given, there a b 20. 3. 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. 91. PROP. xciv. . 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: Be cause each of the points D, A is given, a 29. dat. the straight line A Dis givena in position B b 18. 3. and magnitude: And DBA is a right c Cor.5.4. angle, wherefore DA is a diameter c of the circle DBA, described about the D A А triangle DBA: and that circle is thered 6. def. fore given in position : And the circle d 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. e a 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 в А in position and magnitude : And be cause AD is given, the square of AD c 56. dat. is given “, which is equal to the rect angle BA, AC: Therefore the rectangle BC, AC is given. a 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 cir. cumference 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 C therefore the points E, F are given ; 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 to the c 35. 3. rectangle BA, AC, which consequently is given. a 29. dat. b 28 dat. 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 te 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 d 12. 5. e 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 6 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 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 e 21. 3. to EAC, and the angle ACE to e ADB, the triangle ACE is equian- F A B с 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 5 a Produce CA, and make AF equal to AB, and join BF; S 5. & 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. |