11 Sup. 12 Sup. 13 46, I. 31, I. 143, III. Ax. 3 15 C. 16 C. 17 Rec. EHF touches AB in L; AL2 = EA. AH, i. e. C. D. (AB)=; But if EGGL, and EG > GL; .. EHF cuts AB; on NB desc. NB2 and complete the AP. . LM = LN, and AL= LB,.. AM= NB, .. PN NB; AP= C. D, and AP has been applied to AB, deficient by BN2. Q. E. F. This last Problem may be thus enunciated; "To cut a given line AB in the point N so as to make the rectangle AN. NB equal to a given space." 66 Or, which is the same thing, Having AB the sum of the sides of a rectangle given, and also its magnitude, or area, to find the sides." PROP. 29.-PROB. To a given st line to apply a parallelogram equal to a given rectilineal figure, and exceed ingby a parallelogram similar to another given parallelogram. CON. 10, I. 18, VI. 25, VI. 21, VI. 2, I. 31, I. DEM. 26, VI. 36, I. 43, I. 24; VI. C. 110, I. 18, VI. Bis. AB in E, and on EL descr. & sim. sit. to X; EL sim. GH = EL + Z, & sim. & sim. GH is sim. to EL. Let KH be homol. to FL, & KGʻto FE; GH > EL, .. HK > FL & KG > FE; 6 Pst.2,1.3, 1. Produce FL & FE, & make FLM = & FEN .. KH, KG; MN is sim to EL; MN is eq. & but GH sim to EL, 731, I. 8 C, 6, 7. 9 C. 3. 7s about the same diag. Draw their diag. FX, & complete the scheme, then to AB is applied AX D. 1 C.2, 8. Ax. 1. GH-EL+Z,& GH = ΜΝ 2 Sub. Ax. 3. take away EL; 3 C. 1, 31, I. 43, I. = Z, & ex X. MN, EL+%; .. rem. gnomon NOL= Z. And AE 4 Add. Ax. 2. Add 5 D. 2. 6 Conc. - EB AN=NB — BM. NO, the whole whole gnomon. NOL; AX the N.B.-In the diagram, X by mistake of the engraver occurs twice. SCH. Of the thirteen books of Elements written by Euclid, the tenth bears evident traces of the greatest attention having been bestowed to render it complete. The doctrine of Incommensurables there receives its developement, and is treated with great exactness. "The most conspicuous propositions of elementary geometry," says an eminent writer, "which are applied in the tenth book, are the 27th, 28th, & 29th of the sixth book, of which it may be useful to give the algebraical signification. The first of these (the 27th) amounts to shewing that 2 x- 22 has its greatest value when x 1, and contains a limitation necessary to the conditions of the two which follow. The 28th proposition is a solution of the equation ar x2=b, upon a condition derived from the preceding proposition, namely, that a2 shall exceed b. It might appear more correct to say that the solution of this equation is one particular case of the proposition, namely, where the given parallelogram is a square; but nevertheless, the assertion applies equally to all cases. Euclid, however, did not detect the two solutions of the question; though if the diagonal of a parallelogram in his construction be produced to meet the production of a line which it does not cut, the second solution may be readily obtained. This is a strong presumption against his having anything like algebra; since it is almost impossible to imagine that the propositions of the tenth book, deduced from any algebra, however imperfect, could have been put together without the discovery of the second root. The remaining proposition (the 29th) is equivalent to a solution of ax + x2: b; but the case of 2 axb is wanting, which is another argument against Euclid having known any algebraical reasoning.— PENNY CYC. XII, p. 38. USE and APP.-Several Problems of a like kind to Prop. 29, and in some respects equivalent to it may be here advantageously introduced; PROB. 1°. exscribe To to a given triangle ABC, a parallelogram equal to a given rectilineal figure, AX, and having an angle equal to one of the angles of the given triangle. N.B. This Prob. is equivalent to to the foregoing Prop. 29. Euclid, p. 100. Manual of PROB. 2°. To a given st. line AB to apply a rectangle which shall be equal to a given square, that on C and exceeding by a square.. 4 Sol. D. 1 C. 26, II on BG desc. the sq. BH, and F A then AH = C2, & exceeding by GK is applied to AB. ... AB is divided equally in D and produced to G, 3 Sub. Ax. 3,I. from each take DB2, ... rem. AG. GB=BE2= C2 4 C. 3. 5 6 Conc. but •.• GH = GB,.. AG, GB is rectangle AH; AH = C2, and exceeding by GK, is applied to AB. Q. E. F. N.B.-This Problem is the same as, "To produce a given st. line, AB, so that the rectangle contained by the external segments of the given line may be equal to a given space, as C2." The foregoing Con. and Dem. may be used, and then it will appear that the problem is, to produce AB so that AG. GB = C2. C. 12, I. 11, I. 2 Pst. 1, I. 31,I. 3 Pst. 3, I. Draw AE = C, & BF = D, Ls AB on contrary sides; join EF & bisect EF in G; from G with GE desc. O meeting AE in H; 4 Pst. 1, 31,I. join HF, & draw GL || AE. 5 6 46, I, 31, I. 7 Sol. D. 1 C. 1, 28, I. 5 Ax. 1, I. 6 Conc. Let the in M,N; meet AB produced to AB is applied AP = C.D & exceeding by sq. BP. Q. E. F. N.B. This Problem is the same as,-" To find a point N in a given st. line AB produced, so as to make the rectangle AN. NB a given space." Or, which is the same thing,-" Having given AB the difference of the sides of a rectangle and the magnitude of it, to find the sides." PROP. 30.-PROB. To cut a given st. line in extreme and mean ratio; i. e., so that the whole line shall be to the greater segment as the greater segment to the less. CON. 46, I. To describe a sq. on a given st. line. 11, II. To divide a given st. line into two pts, so that the rect. contained by the whole and one of the pts, shall be equal to the sq. of the other part. DEM. AX. 3, I. 14, VI. 34, I. Def. 30, I. A sq. has all its sides eq, and all its s rt. s. 14. VI. 17, VI. Def. 3, VI. 3 Sol. D. 1 C. 1. CG = □ AD, exceeding by fig. AG, sim. to AD; then AB is cut in extreme & mean ratio in . H. 2 C. 2. Ax.. 3, I. 3 C. 14, VI. .. rem. BK =rem. AG; &. BK is eq, ang. to AG, .. KH: HG = AH : HB; |