and let AC, BD be drawn; the rectangle AC.BD is equal Book VI. to the two rectangles AB.CD, and AD.BC. Make the angle ABE equal to the angle DBC; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC: And the angle BDA is equal a to a 21. 3, the angle BCE, because they are in the same segment; there fore the triangle ABD is equiangular to the triangle BCE. Whereforeb, BC:CE::BD:DA, and confequently c BC.DA= BD.CE. Again, because the angle ABE is equal to the angle DBC, and the angle a BAE to theangle BDC, the triangle ABE is equiangular to the triangle BCD; therefore BA: AE :: BD: DC, and BA.DC=BD.AE; But it was shewn that BC.DA= BD.CE; wherefore BC.DA+BA.DC=BD.CE+BD.AE= BD.AC d. That is the rectangle contained by BD and AD, dr.2. is equal to the rectangles contained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D. PROP. E. THEO R. IF an arch of a circle be bisected, and from the extremities of the arch, and from the point of bisection, straight lines be drawn to any point in the circumference, the sum of the two lines drawn from the extremities of the arch will have to the line. drawn from the point of bisection, the same ratio which the straight line fubtending the arch, has to the straight line fubtending half the arch. AB.CD: but AD.CB+DB.AC=AD.AC+DB.AC, bebr. 2. cause CB AC. Therefore AD.AC+DB.AC, that is b, AD+DB.AC=AB.CD. And because the fides of equal c 14. 6. rectangles are reciprocally proportionalc, AD+DB: DC:: AB: AC. Wherefore, &c. Q. E. D. PROP. F. THEOR. TE F two points be taken in the diameter of a circle, such that the rectangle contained by the segments intercepted between them and the centre of the circle be equal to the square of the radius; and if from these points two straight lines be drawn to any point whatsoever in the circumference of the circle, the ratio of these lines will be the fame with the ratio of the segments intercepted between the ⚫two first-mentioned points, and the circumference of the circle. Let ABC be a circle, of which the centre is D, and in DA produced, let the points E and F be such that the rectangle ED, ED, DF is equal to the square of AD; from E and F to any Book VI. Join BD, and because the rectangle FD, DE is equal to the square of AD, that is, of DB, FD: DB:: DB: DE 2. The a 17.6. two triangles, FDB, BDE have therefore the fides propor fii. g. COR. If AB be drawn, because FB: BE:: FA: AE, the angle FBE is bisected g by AB. Also, fince FD: DC::g3.6. DC: DE, by composition h, FC:DC::CE: ED, and fince it h 18. 5. has been shewn that FA: AD (DC) :: AE: ED, therefore, ex æquo, FA:AE::FC:CE: But FB: BE :: FA: AE, therefore, FB: BE::FC:CEf; so that if FB be produced to G, and if BC be drawn, the angle EBG is bisected by the line BC k; k A. 6. PROP. 1 Book VI: PROP. G. THEOR. IF from the extremity of the diameter of a circle a straight line be drawn in the circle, and if either within the circle or produced without it, it meet a line perpendicular to the fame diameter, the rectangle contained by the straight line drawn in the circle, and the fegment of it, intercepted between the extremity of the diameter and the perpendicular, is equal to the rectangle contained by the diameter, and the segment of it cut off by the perpendicular. Let ABC be a circle, of which AC is a diameter, let DE be perpendicular to the diameter AC, and let AB meet DE in F; the rectangle BA.AF is equal to the rectangle CA.AD. Join BC, and because ABC is an angle in a femicircle, it is a E B F C D A C B A E D F बब a 31. 3. right anglea: Now, the angle ADF is also a right angleb; b Hyp. and the angle BAC is either the fame with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, and C4.6. BA:AC:: AD: AFc; therefore also the rectangle BA.AF, contained by the extremes, is equal to the rectangle AC.AD d 16. 6. contained by the meansd. If therefore, &c. Q. E. D. PROP. T PROP. H. THEOR. HE perpendiculars drawn from the three angles of any triangle to the opposite fides interfect one another in the fame point. Let ABC be a triangle, BD and CE two perpendiculars interfecting one another in F; let AF be joined, and produced if necessary, let it meet BC in G, AG is perpendicular to BC. Join DE, and about the triangle AEF let a circle be described, AEF; then, because AEF is a right angle, the circle described about the angles, the triangles BEF a 31. 3. b 15. 1 C and CDF are equiangular, and therefore BF: EF::CF: FDC, c 4.6. or alternately d, BF: FC:: EF: FD. Since, then, the fides d 6. 5. about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are alfo equiangulare; wherefore the angle e 6. 6. FCB is equal to the angle EDF. But EDF is equal to EAF, because they are angles in the same segment f; therefore the f 21. 3. angle EAF is equal to the angle FCG: Now, the angles AFE, CFG are also equal, because they are vertical angles; therefore the remaining angles AEF, FGC are also equalg: But 8 32. . AEF is a right angle, therefore FGC is a right angle, and AG is perpendicular to BC. Q. E. D. COR し |