PROPOSITION XLIV. PROBLEM. To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given st. line, C the given A, and D the given L. It is required to apply to AB a O=A C, and having one of its Zs= D. Constr. Describe a BEFG = A C, having 2 EBG = LD, (i. 42) so that BE be in the same st. line with AB; produce FG to H, draw AH || BG or EF, (i. 31) and join HB. Dem. Then, : HF falls on the Is AH, EF, .. LS AHF, HFE are together = two rt. LS; (i. 29) i. Ls BHF, HFE are less than two rt. Ls: but st. lines which with another st. line make the two int. Zs on the same side less than two rt. Ls, will meet on that side, if produced far enough: (ax. 12) :: HB, FE shall meet, if produced ; let them meet in K, through K draw KL || EA or FH, and produce HA, GB to meet KL at L, M. = Then KLHF is a o, of which the diameter is HK; and AG, ME are the 7s about HK, also LB, BF are the complements; .. compl. LB = compl. BF; (i. 43) but BF= AC; (constr.).. LB = A C. And :: L. GBE = L ABM, (i. 15) and also = LD; = (constr.) .. L ABM = . D. .. to the st. line AB the LB is applied as required. Q. E. F. PROPOSITION XLV. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, having an angle equal to a given rectilineal angle. Let ABCD be the given rectil. figure, and Ethe given L It is required to describe a O = to ABCD, having an = _ E. Constr. Join DB, and describe O FH = A ADB and having L FKH = LE; (i. 42) to GH apply the OGM = A DBC, having L GHM = L E. (i. 44) Then FKML shall be the o required. Dem. : each of the Zs FKH, GHM = LE, (constr.) .:. FKH = L GHM: add to each of these =S Z KHG ; 1. Zs FKH, KHG= 2s KHG, GHM, (ax. 2) but FKH, KHG together = two rt. Ls; (i. 29) .. KHG, GHM together = two rt. Ls; and : at H in GH the two st. lines KH, HM, on the opposite sides of it, make the adj. Ls KHG, GHM = two rt. Ls, .. HK is in the same st. line with HM. (i. 14) = (i. 30) And :: HG meets the 115 KM, FG, .. L MHG (i. 29) but 2s MHG, HGL = two rt. Zs; (i. 29) .. also HGF, HGL = two rt. <s, and .:. FG is in the same st. line with GL. (i. 14) And : KF is || HG, and HG is || ML, :, KF is || ML: and FL has been proved || KM, .. FKML is a O; (def.) and since o HF = A ABD, and GM = A DBC': :. the whole OKFLM = the whole rectil. figure ABCD. Therefore, the O KFLM has been described as required. Q. E. F. Cor. From this it is manifest how to a given straight line to apply a parallelogram which shall be equal to a given rectilineal figure, and shall have an angle equal to a given rectilineal angle; namely by applying to the given line a parallelogram equal to the first triangle (i. 44), and having an angle equal to the given angle ; and so on. [This appears to be a difficult prop. for beginners, as some step in the demonstration is often omitted. The main point is to show clearly that FGL, KHM are straight lines by means of props. 14 and 29. When the figure FM has thus been proved to be a parallelogram, the rest easily follows.] PROPOSITION XLVI. PROBLEM. To describe a square on a given straight line. AD= (i. 34) Constr. From A draw AC I AB; (i. 11) make AD = AB, (i. 3) through D draw DE || AB, and through B, BE | AD; (i. 31) Then shall ADEB be a sq. Dem. Since AE is a (constr.) .. AB = DE and BE; but AD= AB, (constr.) .. AB, BE, ED, DA are all =, and the figure AE is equilateral. Again, since AD meets the || AB, DE 1. s BAD, ADE are together = two rt. Zs; (i. 29) but BAD is a rt. L ; (constr.).. also ADE is a rt. L. Now the opp. Zs of os are =; (i. 34)each of the opp. LS ABE, BED is a rt. L; .. AE is rectangular, and it has been proved equilateral; :: it is a sq: (def. 30) and it is described on the given st. line AB. Q. E, F. Cor. Hence every that has one rt. L has all its ZS rt. Ls. |