PROPOSITION D. THEOREM. The rectangle, contained by the diagonals of a quadrilateral inscribed in a circle, is equal to the sum of the rectangles, contained by its opposite sides. B D Let ABCD be any quadrilateral inscribed in a . Join AC, BD. Then rect. AC, BD=rect. AB, CD together with rect. AD, BC. Make ABE: = 4 DBC; I. 23. .. AD is to BD as CE is to BC, and add to each the ▲ EBD. and ▲ BDA = 4 BCE in the same segment; III. 21. ..▲ ABD is equiangular to ▲ BCE, I. 32. VI. 4. VI. 16. III. 21. I. 32. VI. 4. VI. 16. and.. rect. AD, BC=rect. BD, CE. 4 BDC, in the same segment, ..▲ ABE is equiangular to ▲ BCD. Δ .. AB is to AE as BD is to CD, and.. rect. AB, CD=rect. BD, AE. Hence rect. AB, CD together with rect. AD, BC =rect. BD, AE together with rect. BD, CE. II. 1. Q. E. D. Ex. If the diagonals cut one another at an angle equal to one third of a right angle, the rectangles contained by the opposite sides are together equal to four times the quadrilateral figure. PROPOSITION XXVI. THEOREM. (Eucl. VI. 23.) Equiangular parallelograms have to one another the ratio, which is compounded of the ratios of their sides. Let AC and CF be equiangulars, having ▲ BCD = ▲ ECG. Then must AC have to CF the ratio compounded of the ratios of their sides. Let BC and CG be placed in a straight line. Then DC and CE are also in a straight line. Complete the DG, and taking any st. line K, I. 14. VI. 11. and make as DC is to CE so L to M. VI. 11. make as BC is to CG so K to L Then K has to M the ratio compounded of the ratios of K to L and L to M, the sides. .. K has to M the ratio compounded of the ratios of VI. Def. 3, p. 260. PROPOSITION XXVII. THEOREM. (Eucl. VI. 24). Parallelograms about the diameter of any parallelogram are similar to the whole parallelogram and to one another. Let ABCD be a, of which the diameter is AC; and .. AB is to BC as AE is to EF; and since the opposite sides of the Os are equal, .. AB is to AD as AE is to AG, VI. 4. V. 6. and DC is to CB as GF is to FE, V. 6. and CD is to DA as FG is to GA. V. 6. Thus the sides of the Os AEFG, ABCD about their equal angles are proportional. Ex. Show that each of the complements of the parallelogram is a mean proportional between the parallelograms about the diameter. PROPOSITION XXVIII. THEOREM. (Eucl. VI. 26.) If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter. Let the B s ABCD, AEFG be similar and similarly situated, and have DAB common. Then must ABCD and AEFG be about the same diameter. For, if not, let ABCD have its diameter, AHC, not in the same st. line with AF, the diameter of AEFG. I. 31. Let GF meet AHC in H, and draw HK || to AD. the less-the greater, which is impossible. V. 5. V. 8. .. ABCD and AKHG are not about the same diameter, and.. ABCD and AEFG must have their diameters in the same st. line, that is, they are about the same diameter. Q. E. D. PROPOSITION XXIX. PROBLEM. (Eucl. VI. 25.) To describe a rectilinear figure which shall be similar to one, and equal to another, given rectilinear figure. B од E M Let ABC and D be two given rectilinear figures. It is required to describe a figure similar to ABC and equal to D. On BC describe the on CE describe the BLEC equal to ABC, and I. 45, Cor. and having FCE = 4 CBL. Then BC and CF are in a straight line, and LE and EM are in a straight line. I. 45, Cor. I. 29 and 14. Find GH, a mean proportional between BC and CF, VI. 13. and on GH describe the rectilinear figure KGH, similar and similarly situated to ABC. Then BC is to GH as GH is to CF, VI. 18. .. as BC is to CF so is ABC to KGH. VI. 20, Cor. 2. VI. 1. and .. as ABC is to KGH so is BE to EF. V. 5. Now ABC is equal to BE, and.. KGH =□ EF. But EF-the figure D. .. KGH =D; and KGH is similar to ABC. Hence a figure KGH has been described as was required. Q. E. F. Constr. V. 14. H |