PROP. XXIV. THEOR. 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 parallelograms about the diameter: the parallelograms EG, HK shall be similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the angle ADC is equal to the angle AGF: and because BC, EF *29. 1. are parallels, the angle ABC is equal to the angle AEF; and each of the angles BCD, EFG is equal *34. 1. to the opposite angle DAB, and therefore they 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* AB: BC:: AE *4. 6. : EF; or, because the opposite #34. 1. sides of parallelograms are equal to one another,* E B H CD. DA :: FG : GA; therefore the sides of the parallelograms ABCD, AEFG about the equal angles D are proportionals; and they are therefore +1 Def. 6. similar to one another; for a like reason, the parallelogram ABCD is similar to the parallelogram FHCK; wherefore each of the parallelograms GE, KH is similar to DB; but rectilineal figures which are similar to the same rectilineal figure are also simi#21. 6. lar to one another: therefore the parallelogram GE is similar to KH. Wherefore parallelograms, &c. Q. E. D. PROP. XXV. PROB. To describe a rectilineal figure which shall be similar to one, and equal to one another given rectilineal figure. Let P be the given rectilineal figure to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar to P, and equal to D. #45. #45. 1 Cor. Upon the straight line BC describe* the parallel1 Cor. ogram BE equal to the figure P; also upon CE describe the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL; therefore, LBC, ECB being equal to two right angles FCE, ECB are equal to two right angles, and BC and CF are in a straight line, as also, for similar reasons, are LE and EM. Between BC and CF find a mean proportional GH, and upon GH describe the rectilineal figure Q, similar and similarly situated to the figure P. #14. 1. *13. 6. #18. 6. *20. 6. 2 Cor. #2. 5. * Because BC: GH :: GH: CF, and that if three straight lines be proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure upon the second; therefore, BC: CF :: P:Q; but BC: CF :: BE : EF; therefore* P Q :: BE: EF; and the Cor. 1. +Const. figure P is equal to the parallelogram BE; there#9. 5. fore the figure Q is equal to the parallelogram +Const. EF; but EF is equal to the figure D; wherefore also Q is equal to D: and it is similar to P. Therefore the rectilineal figure Q has been described similar to the figure P, and equal to D. Which was to be done. PROP. XXVI. THEOR. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter. D Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common: ABCD and AEFG shall be about the same diameter. For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the parallelogram EG, and let GF, or GF prolonged, meet AHC in H; and through H draw HK parallel to AD or BC: therefore the parallelograms B ABCD. AKHG being about the same #24.6. +Hyp #2. 5. #9. 5. K H E diameter, are similar to one another: wherefore DA: AB GA: AK; but because ABCD and AEFG are similart parallelograms, DA: AB :: GA: AE; therefore* GA: AE :: GA: AK; and the antecedents being equal, the consequents are equal; therefore AK is equal to AE, which is impossible, since one must be less than the other; therefore ABCD and AKHG are not about the same diameter: that is, the diameter of BD cannot be in a different straight line from AF; wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c. Q. E. D. PROP. XXVII. THEOR. Of all the rectangles contained by the segments of a given straight line, the greatest is the square described upon half the line. Let AB be a given straight line, which is bisected in C, and let D be any other point in it. The square on AC is greater than the rectangle AD · DB. For, since the straight line AB C D + + B is divided into two equal parts in A. C, and into two unequal parts in D, the rectangle AD DB together with the square of CD, is equal to the square of half the line; and, consequently, the square of half the line must be greater than the rectangle AD DB alone. Therefore, of all the rectangles, &c. Q. E. D. PROP. XXVIII. PROB. To divide a given straight line so that the rectangle contained by its segments may be equal to a given square, not greater than the square of half the line. Let AB be the given line, and C the side of the given square. #31. 1. AD. With the centre D at the distance DF, describe a circle which must pass through the points A, B, because DA, DF, DB are, by construction, equal. Draw EP parallel to AB* meeting the circle in P, and draw PG parallel to ED, meeting AB in G; then G will be the point of division required. #34. 1. For, by construction, EG is a rectangle; therefore PG is equal to ED; but ED was made equal to C; therefore PG is equal to C. But the rect#13. 6. angle AG GB is equal to the square of PG;* therefore the rectangle AG GB is equal to the square of C; and thus AB is divided in G as required. PROP. XXIX. PROB. To prolong a given straight line so that the rectangle contained by the segments between the extremities of the given line and the point to which it is prolonged may be equal to a given square. Let AB be the given line; and let C be the side of the given square. Bisect AB in D and draw BE at right angles to it, so that BE may be equal to C; and, having joined D, E, from the centre D, at the distance DE describe a circle meeting AB, prolonged in G and F. Because D is the centre of the circle, DF F is equal to DG; also DA is equal to DB by construction; hence the remainders AF, BG are equal. To each of these add AB, then FB is equal to AG. But the rectangle FB BG is equal to the square of A G BE; therefore the rectangle AG BG is equal to #13. 6. the square of BE, or to the square of C; AB, therefore, is produced to G, as required. NOTE. This, as well as the two propositions immediately preceding, we have, after the example of Playfair, substituted for those of Euclid, which, though somewhat more general, are remarkably operose and inelegant. The demonstrations given above of the 28th and 29th, will be found to be simpler and shorter than those of Playfair. |