A DH G B C Let BC, CG, be placed in a straight line; therefore DC and CE are also in a straight line (2. 1.); and complete the parallelogram DG; and, taking any straight line K, make (12. 5.) as BC to CG, so K to L: and as DC to CE, so make (12.5.) L to M: therefore the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE. But the ratio of K to Mis that which is said to be compounded (15. Def. 4.) of the ratios of K to L, and L to M; Wherefore also Khas to M the ratio compounded of the ratios of the sides; and because as BC to CG, so is the parallelogram AC to the parallelogram CH (2. 5.); but as BC to CG, so is K to L; therefore K is (15. 4.) to L, as the parallelogram AC to the parallelogram GH: KLM EF again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but as DC to CE, so is L to M; wherefore Lis (15. 4.) to M, as the parallelogram CH to the parallelogram CF: therefore, since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; (16.4.) K is to M, as the parallelogram AC to the parallelogram CF: but K has to M the ratio which is compounded of the ratios of the sides: therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q. E. D. PROP. XXIV. THEOR. The parallelograms about the diameter of any parallelogram are similar to the whole, and to one another. Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter: the parallelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the angle ADC is equal (3.2.) to the angle AGF: for the same reason, because BC, EF are parallels, the angle ABC is equal to the angle AEF; and each of the angles BCD, EFG is equal to the opposite angle DAB (9. 2.), and therefore are equal to one another; wherefore the parallelograms ABCD, AEFG are equiangular; and because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore (5. Def. 5.) as AB to BC, so is AE to EF: and because the opposite sides of parallelograms are equal to one another (9. 2.), AB is (13. 4.) to AD, similar to DB: but rectilineal figures which are similar to the same rectilineal figure, are also similar to one another (21. 5.), therefore the parallelogram GE is similar to KH. Wherefore the parallelograms, &c. Q. E. D. PROP. XXV. PROB. To cut a given straight line in extreme and mean ratio. Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC (24. 2.); then because the rectangle AB, BC is C equal to the square of AC, as BA to AC, so is AC to CB (17.5.): therefore AB is cut in extreme and mean ratio in C (2. Def. 5.). Which was to be done. PROP. XXVI. THEOR. In right angled triangles the rectilineal figure described upon the side opposite to the right angle, is equal to the similar, and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC; the rectilineal figure described upon BC is equal to the similar and similarly described figures upon BA, AC. Draw the perpendicular AD; therefore because in the right an gled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one another: and because the triangle ABC is similar to ADB, as CB to BA, so is BA to BD (5.5.); and because these three straight lines are proportionals, as the first to the third, so is the figure upon the B i D C first to the similar, and similarly described figure upon the second (2. Cor. 20. 5.): therefore as CB to BD, so is the figure upon CD to the similar and similarly described figure upon BA: and, inversely (5. 4.), as DB to BC, so is the figure upon BA to that upon BC; for the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore as BD and DC together to BC, so are the figures upon BA, AC to that upon BC (18.4.): but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D. PROP. XXVII. THEOR. In equal circles, angles at the centres have the same ratio to one another as the circumferences on which they stand. Let ABCD and FGHI be equal circles, and let the angles AEB and FKG be at the centre; they have the same ratio to each other as the circumferences on which they stand, AB and FG. Produce AE, BE to the circumference at D and C; and FK, GK to the circumference at H and I; then the angles at E are together 1 equal to four right angles (1. Cor. 3. 1.), and they stand on the whole circumference; and therefore, whatever part the angle AEB is of four right angles, the arc AB is of the whole circumference; and therefore, as the angle AEB is to four right angles, so is the arc AB to the whole circumference ABCD (18. Def. 4.). It may be shewn in the same way, that as the angle FKG is to four right angles, so is the arc FG to the whole circumference FGHI; but the whole circumference FGHI is equal to the whole circumference ABCD by the hypothesis: and therefore, as the angle AEB is to the angle FKG, so is the arc AB to the arc FG. Therefore, in equal circles, &c. Q. E. D. Cor. I. It may be shewn in the same way, that in equal circles the sectors have the same ratio to each other as the circumferences on which they stand. COR. II. Because the angle at the centre is double the angle at the circumference upon the same base; therefore, in equal circles angles at the circumference have the same ratio to one another as the circumferences on which they stand. PROP. XXVIII. THEOR. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD and by 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 (13. 3.) to the angle BCE, be B cause they are in the same segment; and the angle (13. 3.) BAE to the an gle BDC, the triangle ABE is equian C gular to the triangle BCD: as therefore BA to AE, so is BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE: but the rectangle BC, AD has been shewn equal to the rectangle BD, CE; therefore the whole rectangle AC, BD is equal to the rectangle AB, DC together with the rectangle AD, BC. Therefore, the rectangle, &c. Q. E. D. THE ELEMENT OF GEOMETRY. BOOK VI. DEFINITIONS. I. WHEN all the angles of a plane rectilineal figure are upon the circumference of a circle, the figure may be said to be inscribed in the circle, or the circle circumscribed about the figure. II. When each side of a figure touches the circumference of a circle, the figure may be said to be circumscribed, or the circle inscribed. III. A plane rectilineal equilateral and equiangular figure may be called a regular polygon. IV. The sum of the sides of a figure may be called the perimeter of the figure. V. When the extremities of a straight line are in the circumference of a circle, the line may be said to be placed in the circle. |