pendicular 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 semicircle, it is a right angle (31.3.): Now, the angle ADF is also a right angle (Hyp.); and the angle BAC is either the same with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, and BA: AC :: AD: AF (4.6.); therefore also the rectangle BA.AF, contained by the extremes, is equal to the rectangle AC.AD contained by the means (16.6.). If therefore, &c. Q. E. D. 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 have AF for its diameter (31. 3.). In the E A D same manner, the circle described about the triangle ADF has AF for its diameter; therefore the points A, E, F and Dare in the circumference of the same circle. But because the angle EFB is equal to the angle DFC (15. 1.), and also the angle BEF to the angle CDF, being both right angles, the triangles BEF and CDF are equiangular, and therefore BF: EF :: CF : FD (4.6.), or alternately (16.5.) BF: FC :: EF: FD. Since, then, the sides about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are also equiangular (6.6.); wherefore the angle FCB is equal to the angle EDF. But EDF is equal to EAF, because they are angles in the same segment (21.3.); therefore the 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 equal (32. 1): But AEF is a right angle, therefore FGC is a right angle, and AG is perpendicular to BC. Q. E. D. G C COR. The triangle ADE is similar to the triangle ABC. For the two triangles BAD, CAE having the angles at D and E right angles, and the angle at A common, are equiangular, and therefore BA: AD:: CA: AE, and alternately BA: CA:: AD: AE; therefore the two triangles BAC, DAE, have the angle at A common, and the sides about that angle proportionals, therefore they are equiangular (6. 6.) and similar. Hence the rectangles BA.AE, CA.AD are equal PROP. K. THEOR. If from any angle of a triangle aperpendicular be drawn to the opposite side or base; the rectangle contained by the sum and difference of the other two sides, is equal to the rectangle contained by the sum and difference of the segments, into which the base is divided by the perpendicular. Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, so that BD, DC are the segments of the base; (AC+AB) (AC-AB)=(CD+DB) (CD-DB). From A as a centre with the radius AC, the greater of the two sides, describe the circle CFG: produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF=AC, BF=AB+AC, the sum of the sides; and since AE=AC, BE-AC, -AB= the difference of the sides. Also, because AD drawn from the centre cuts GC at right angles, it bisects it; therefore, when the perpendicular falls within the triangle, BG=DG-DB=DC-DB= the difference of the segments of the base, and BC=BD+DC= the sum of the segments. But when AD falls without the triangle, BG= DG+DB=CD+DB= the sum of the segments of the base, and BC =CD-DB = the difference of the segments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BE=CB BG; that is, as has been shown, (AC+AB) (AC-AB) = (CD+DB) (CD-DB). Therefore, &c. OF GEOMETRY. SUPPLEMENT. BOOK I. OF THE QUADRATURE OF THE CIRCLE A arch. DEFINITIONS. I. CHORD of an arch of a circle is the straight line joining the extremities of the arch; or the straight line which subtends the II. The perimeter of any figure is the length of the line or lines, by which it is bounded. III. The area of any figure is the space contained within it. AXIOM. The least line that can be drawn between two points, is a straight line: and if two figures have the same straight line for their base, that which is contained within the other, if its bounding line or lines be not any where convex toward the base, has the least perimeter. COR. 1. Hence the perimeter of any polygon inscribed in a circle, is less than the circumference of the circle. COR. 2. If from a point two straight lines be drawn touching a circle, these two lines are together greater than the arch intercepted between them; and hence the perimeter of any polygon described about a circle is greater than the circumference of the circle. |