Wherefore, upon the given straight line AB the segment AHB of a circle is described which contains an angle equal to the given angle at C. Which was to be done, PROP. XXXIV. PROB. To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D. Draw (17. 3.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF make (23. 1.) the angle FBC equal to the angle D; therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (32. 3.) to the angle in the alternate segment BAC but the angle FBC is equal to the angle D; therefore the angle in the : D E B segment BAC is equal to the angle D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC is equal. to the rectangle contained by BE, ED. If AC, BD pass each of them through the centre, so that E is the centre, it is evident, that AE, EC, BE, ED, being all equal, the rectangle AE. EC is likewise equal to the rectangle BE.ED. But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles in the A E D B C point E: then, if BD be bisected in F, F is the centre of the circle M ABCD; join AF and because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre at right angles in E, AE, EC are equal (3. 3.) to one another: and because the straight line BD is cut into two equal parts in the point F, and into two unequal, in the point E, BE.ED (5. 2.)+ EF2 FB2= AF2. But AF2: AE+ (47. 1.) EF2, therefore BE.ED + EF2 = AE2+EF2, and taking EF2 from each, BE. ED AE AE.EC. F A C E B D Next, Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (12. 1.) FG perpendicular to AC thereis fore AG is equal (3. 3. to Gerefore AE.EC+(5. 2EGAGs, and adding GF to both, AFEC+EG2+GF2=AG2 +GF2. Now G+ GF2 = EF2, and F E Ꮐ B But FB* AG2+GF2=A therefore AE.EC+EF2=AF2=FB2. 'BE.ED+ (5. 2. EF2, therefore AE.EC+EF2=BE.ED+EF2, and taking EF2 from both, AE.EC.-BE.ED. Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEH and because, as has been shown, AE.EC : = GE.EH, and BE.ED = GE.EH; therefore AE.EC BE.ED. Wherefore, if two straight lines, &c. Q. E. D. H F E A If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the square of the line which touches it. Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle, and DB touches it: the rectangle AD.DC is equal to the square of DB. Either DCA passes through the centre, or it does not; first, Let it pass through the centre E, and join EB; therefore the angle EBD is a right (18. 3.) angle and because the straight line AC is bisected in E, and produced to the point D, AD.DC + EC2 ED2 (6.2.). But EC - EB, therefore AD. B DC + EB2 = ED2. Now ED2 = (47. 1.) EB2 + BD2, because EBD is a right angle; therefore AD.DC+EB EB2 + BD2, and taking EB2 from each, AD.DC BD2. But, if DCA does not pass through the centre of the circle ABC, take (1. 3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED: and be cause the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it likewise bisects (3. 3.) it; therefore AF is equal to FC; and because the straight line AC is bisected in F, ⚫ and produced to D (6. 2.), AD.DC+FC B FD2; add FE2 to both, then AD.DC+FC2 +FE2FD2+FE2. But(17. 1.)EC2=FC2 +FE2, and ED = FD3 FÉ2, because DFE is a right angle, therefore AD.DC+ EC-ED2. Now, because EBD is a right angle, EDEB2+BD2=EC2+BD2, and therefore, AD.DC+EC2 EC2+BD2, and AD.DC BD2. A Wherefore, if from any point, &c. Q. E. D. COR. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and he parts of them without the circle, are equal to one another, viz. BA.AE=CAAE for each of these rectangles is equal to the square of the straight line AD, which touches the circle. F B PROP. XXXVII. THEOR. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line, which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets shall touch the circle. Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD.DC, be equal to the square of DB, DB touches the circle. Ꭰ Draw (17. 3.) the straight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right (18. 3.) angle and because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal (36. 3.) to the square of DE; but the rectangle AD.DC is, by hypothesis, equal to the square of DB : therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB: but FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal (8. 1.) to the angle DBF and DEF is a right angle, therefore also LBF is a right angle: but FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (16. 3.) the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. B F E A ELEMENTS OF GEOMETRY. A BOOK IV. DEFINITIONS. I. RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. II. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. |