Book VI. a 21. 3. b 4. 6. 16. 6. d 2. Make the angle ABE equal to the angle DBC; dď to each of these the common angle EBD, then the angle ABD is equal to the angle EBC: And the angle BDA is equal to to the angle BCE, because they are in the same segment; there fore the triangle ABD is equi- BC CE:: BD: DA, and consequently a B BD: DC, and BA.DCBD.AE: But it was shewn that BC.DA=BD.CE; wherefore BC.DA+BA.DC= BD.CE+BD.AE=BD.AC. That is the rectangle contained by BD and AC is equal to the rectangles contained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D. > PROP. E. THEOR. 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 subtending the arch has to the straight line subtending half the arch. Let ABD be a circle, of which AB is an arch bisected in C, and from A, C, and B to D, any point whatever in the circumference, let AD, CD, BD be drawn; the sum of the two lines AD and DB has to DC the same ratio Book VI. that BA has to AC. A B For since ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, AD.CB+ DB.ACAB.CD: but AD.CB+DB.AC=AD.AC + a D. 6. DB.AC, because CB-AC. Therefore AD.AC+DB.AC, that is, (AD+DB) ACAB.CD. And because the b 1.2. sides of equal rectangles are reciprocally proportional o, c 14. 6. AD+DB: DC:: AB: AC. Wherefore, &c. Q. E. D. PROP. F. THEOR. If two points be taken in the diameter of a circle, such that the rectangle contained by the seg ments 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 same 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, DF is equal to the square of AD; from E and F to any point B in the circumference, let EB, FB be drawn; FB: BE:: FA : AE. Book VI. a 17. 6. Join BD, and because the rectangle FD, DE is equal to the square of AD, that is, of DB, FD: DB:: DB: DE, b 6. 6. c 4. 6. d 16. 5. e 17. 5. f 11. 5. g 3. 6. The two triangles, FDB, BDE have therefore the sides proportional that are about the common angle D; therefore they are equiangular, the angle DEB being equal to the angle DBF, and DBE to DFB. Now, since the sides about these equal angles are also proportional, FB: BD :: BE: ED, and alternately, FB: BE :: BD: ED, or FB: BE::AD: DE. But because FD: DA:: DA: DE, by division, FA: DA:: AE: ED, and alternately, FA: AE: :DA: ED. Now, it has been shewn that FB: BE:: AD: DE, therefore FB: BE:: FA: AE. Therefore, &c. Q. E. D. g COR. If AB be drawn, because FB: BE:: FA: AE, the angle FBE is bisected by AB. Also, since FD: DC h 18. 5. DC: DE, by composition ", FC: DC : : CE : ED, and since it has been shewn that FA : AD (DC) : : AE: ED, therefore, ex æquo, FA: AE:: FC: CÉ. FB: BE:: FA: AE, therefore, FB: BE:: FC: CE; so that if FB be produced to G, and if BC be drawn, kA. 6. the angle EBG is bisected by the line BC k. But 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 same diameter, the rectangle contained by the straight line drawn in the circle, and the segment 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 the 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 Book VI. in a semicircle, it is a right angle: Now, the angle ADF a 31. 3. is also a right angle"; and the angle BAC is either the b Hyp. same with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, and BA: AC:: AD: AF; c 4. 6. therefore also the rectangle BA.AF, contained by the extremes, is equal to the rectangle AC.AD contained by the means. If therefore, &c. Q. E. D d 16. 6. Book VI. PROP. H. THEOR. The perpendiculars drawn from the three angles of any triangle to the opposite sides intersect one another in the same point. Let ABC be a triangle, BD and CE two perpendiculars intersecting 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 triangle AEF will a 31. 3. tera. In the same man- angle EFB is equal to b 15. 1. the angle DFC, and also the angle BEF to c 4. 6. € 6. 6. B E the angle CDF, being both right angles, the triangles BEF and CDF are equiangular, and therefore BF: EF CF: FD, or alternatelyd BF: FC:: EF: FD. Since, then, the sides about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are also equiangular; wherefore the angle FCB is equal to the angle EDF. But EDF is equal to EAF, because they are f21. 3. angles in the same segment f; therefore the angle EAF is equal to the angle FCG: Now, the angles AFE, CFG are also equal, because they are vertical angles; therefore 32. 1. the remaining angles AEF, FGC are also equal: But AEF is a right angle, therefore FGC is a right angle, and AG is perpendicular to BC. Q. E. D. COR. The triangle ADE is similar to the triangle ABC. For the two triangles BAD, CAE having the |