E, H, and F, I and G, K: The straight lines EH, FI, and GK, will divide BD into the required number of equal parts. For, in BC, and DA, take KC, and EA, each equal to BH, and join A, B and D, C: Then, since (constr.) AD is equal and parallel to BC, AB is also (E. 38. 1.) parallel to DC; ... ABCD is a , of which BD is a diameter; (constr. and S. 48. 1.) BD is divided by EH, FI, and GK, into as many equal parts as BC, or AD, is divided into. Otherwise, Draw BC, as before, and make the number of equal parts BH, HI, IK, KC, equal to the given number into which BD is to be divided; join C, D; and draw HL, IM, KN, each parallel to CD: Then will these parallels divide BD into the required number of equal parts. For, if LP, MQ, NR be drawn each parallel to BC, it may be proved, (as in S. 48. 1.) that BL LMMNND. PROP. L. 67. PROBLEM. Upon a given finite straight line to describe an equilateral and equiangular octagon. Let AB be a given finite straight line: Upon AB, it is required to describe an equilateral and equiangular octagon. From the points A and B draw (E. 11.1.) AD and BE to AB, and produce, AB both ways to X and Y; bisect (E. 9.1.) the DAX, EBY, by AG, and BL, and make AG and BL each equal to AB; from the points G and L, draw GF to AG, and GH parallel to AD, and make GH AB or AG; in like LI to BL; also, draw (E. 31. 1.) manner, draw LK parallel to BE and make LK= AB; lastly, draw HD parallel to GF, meeting AD in D, and KE parallel to LI, meeting BE in E; and join D, E: The figure ABLKEDHG, described on AB, is an equilateral and equiangular octagon. For, since (constr.) the side AG, of the A AGF, is equal to the side BL, of the A BLI, and GAFLBI, and that the BGF, BLF are equal, being right ; ..., (E. 26. 1.). AF BI; ⇓ also, since (constr.) HF and KI are , FD=GH, and IE = LK; but (constr.) GH = LK; .'. FD=IE; ... the whole AD=the whole BE; and (constr. and E. 28. 1.) AD is parallel to BE; ... (E. 33. 1.) DE is equal and parallel to AB; and .. (constr. and E. 34. 1.) the ADE, BED are right Again, since (constr.) the AGF, BLI, are right, and that the GAF, IBL are each the half of a right,.. (E. 32. 1.) the GFA, LIB, are each the half of a right ▲ ; .'. AG= GF=BL=LI; and (E. 34. 1.) HD=GF, and KE=LI; whence it is manifest that the figure ABLKEDHG is equilateral. Lastly, since HG is parallel to DA, and KL to EB, and FG and IL meet these parallels, .. (E. 29. 1.) the HGF=▲ GFA, and ... HGF= the half of a right ▲; .. (E. 34. 1.) the HDF is the half of a right ; in the same manner, it may be shewn that each of the IEK, KLI, is the half of a right. ; and it has been proved that the ADE, BED are right; whence, and from the construction, it is manifest, that the figure ABLKEDHG, which has been shewn to be equilateral, is also equiangular. PROP. LI. 68. THEOREM. If either diameter of a parallelogram be equal to a side of the figure, the other diameter shall be greater than any side of the figure. Let the diameter AB, of the ACBD, be equal to the side AC: The other diameter CD shall be greater than either AC or AD. For, because AC AB, the ▲ ACB ABC (E. 5. 1.) and (hyp. and E. 29. 1.) the ▲ DAB= < ABC; but the DAC > ▲ DAB; .. the 2 DAC > < ACB; and the sides DA, AC, of the A DAC, are (E. 34. 1.) equal to the sides BC, CA, of the ▲ BCA; .. (E. 24. 1.) CD> AB; but (hyp.) ABAC; .. CD> AC: And it has been shewn that the DAC>ACB; much more then is the 19. 1.) DC > AD. DAC> < ACD; .. (E. PROP. LII. 69. PROBLEM. From a given point to draw a straight line cutting two parallel straight lines, so that the part of it, intercepted between them, shall be equal to a given finite straight line, not less than the perpendicular distance of the two paral lels. Let A be a given point; XY and ZW two given parallel straight lines, indefinite in length; and B a given finite straight line, not less than the perpendicular distance of XY from ZW: It is required to draw through A, a straight line, cutting XY and ZW, so that the part of it, between the two parallels, shall be equal to B. Take any point C in ZW; from C as a centre, at a distance equal to B, describe a circle, cutting in XY in D; join C, D; and through A draw (E. 34. 1.) AF parallel to DC, cutting XY and ZW in the points E and F: Then is EF= B. For, (constr.) the figure DCFE is a □; .·. (E. 34. 1.) EF = DC; and (constr.) DC=B; .. EF-B. PROP. LIII. 70. THEOREM. If, from the summit of the right angle of a scalene right-angled triangle, two straight lines be drawn, one perpendicular to the |