DEM. 1. by Hyp.1.. ABCD is a , and AC its diameter; 2 P. 34. 3 Hyp. 2. 4 P. 34. 5 P. 34. 6 D. 4 & 5. 7 Ax. 2. 8 D. 2. 9 Ax. 3. 10 Recap. But the whole ▲ ABC the whole ▲ ADC; therefore the rem. complement BK complement KD. = the rem. Wherefore, the complements of the parallelograms, &c. Q.E.D. SCH.--The parallelograms about the diagonal, and also their complements, have each an angle in common with the whole parallelogram, and therefore are equiangular with it. USE AND APP.-If any parallelogram, as KD in the last figure, be given, another parallelogram, KB, may be found, equal to it, and having one side, EK, equal to a given line. For, produce FK, and from K set off a distance equal to the given line EK; produce the sides DH, DF, and HK indefinitely; and through E draw AB parallel to HG or DC; draw the diagonal A K, and produce it until it cuts DF produced in C; through C draw a parallel to KF or HD, and it will cut HK and AE produced in the points G and B-then the parallelogram BK will be equal to the given parallelogram KD. PROP. 44.-PROB. To a given st. line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle. SOL.-P. 42. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. P. 31. To draw a st. line, through a point, parallel to a given st. line. Pst. 2. A terininated st. line may be produced to any length in a st. line. DEM.-P. 29. If a st. line fall upon two parallel st. lines, it makes the alternate angles equal, and the ext. ang. equal to the int. and opposite angles, and likewise the two interior angles upon the same side equal to two rt. angles. Ax. 9. The whole is greater than its part. Ax. 12. A st. line meeting two st. lines, so as to make the two int. angles on the same side less than two rt. angles, these two st. lines, being produced, shall meet on the side on which the angles are less than two rt. angles. P. 43. The complements of the parallelograms which are about the diameter of any parallelogram, are equal. Ax. 1. Magnitudes equal to the same, are equal to one another. P. 15 If two st. lines cut one another, the vertical angles are equal. line with AB; 3 Pst. 2. 4 P. 31,Pst.1. 5 Pst. 2. produce FG to. H; and through. A draw AH || BG or EF, Next produce HB and FE to meet in the 6 P.31,Pst. 2. through K draw KL || AE; 7 Sol. and produce GB and HA to M and L; then the BL = the ▲ C, DEM. 1 by C. 3 & 4. . st. line HF falls on the parallels A Hand EF, 2 P. 29. 3 Ax. 9. .. the s AHF and HFE together = two rt. angles; and s BHF and HFE together< two rt. ≤ s; DEM. 4 Ax. 12. 5 Concl. 6 Pst. 2. 8 Pst. 2. 9 Concl. but when int. angles on the same side are <two rt. angles, their sides meet, if produced; .. HB and FE, being produced, will meet: Let them be produced and meet in the point K ; through K draw KL || EA or FH; and produce HA to. L, and GB to. M; then FKLH is a, HK the diameter, A G and MEs about HK, and LB and FB the complements; and .. compl. LB = compl. BF: But And. 10 P. 43. 11 C.1 & Ax. 1. .. also 14 Recap. BF = ▲ C, :. — LB = ▲ C. Therefore, to the st. line AB the parallelogram, &c. Q. E. F. SCH.-1. In this Problem the Solution and the Demonstration are almost unavoidably mixed together. 2. When a parallelogram is drawn on a straight line, it is said to be applied to that line. USE AND APP. This proposition contains a kind of Geometrical Division. The whole area of one figure, as C, or BF, in the last diagram, being given, and AB the side of another; what is required is, so to separate the parts of the given figure, that when applied to the given line, the same number of parts shall be contained in the required figure, as existed in the given figure. Arithmetically we can say, a given triangle contains, for example, 20 square feet—a given line, 4 lineal feet; how can we apply a parallelogram to the 4 lineal feet, so that the area of the parallelogram shall equal that of the triangle? We divide 20 by 4, and the quotient 5 is the lineal measurement of the other side of the parallelogram. Geometrically we say, a given rectangle BC contains 12 square feet—a given st. line BD, or its equal EH, 2 lineal feet; how can we apply another rectangle to BD or EH measuring 2 lineal feet, so that the area of the required rectangle may equal the area of the given rectangle? For, complete the rectangle ADGF; EH equals BD, and EI equals CF; and the complement EG is equal to the complement EA: and if we divide AE into equal squares, and EG into equal squares, we find that in the first there are 12, and in the second 2 × 6, or 12 also. PROP. 45.-PROB. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. SOL.-Pst. 1. A st. line may be drawn to join any two points. P. 42. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. P. 44. To a given st. line to apply a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle. DEM.-AX. 1. Magnitudes which are equal to the same, are equal to each other. Ax. 2. If equals be added to equals, the wholes are equal. P. 29. If a st. line fall on two parallel st. lines, it makes the alternate angles equal; and the ext. angle equal to the int. and opp. angle on the same side; and the two int. angles on the same side equal to two rt. angles. P. 14. If at a point in a st. line two other st. lines on opposite sides make the adj. angles equal to two rt. angles, the two st. lines are in one and the same st. line. P. 30. St. lines parallel to the same st. line, are parallel to each other. Def. A. A parallelogram is a four-sided figure of which the opposite sides are equal and parallel, and the diagonals join opposite angles. 2 Ax. 1. 4 Ax. 2. = LFKH, and also :: FKH = LGHM: KHG; To each of the equal angles add the then thes FKH+ KHG = Ls GHM + KHG: 8 P. 14. 9 C. 9. 10 P. 2. alt. HGF; 11 Add. 12 Ax. 2. = Thus at. H in st. line HG, the adj. s GHM, GHK two rt. angles : .. KH is in the same st. line with HM. = To each of these equal angles add thens MHG and HGL = LS HGF and HGL: HGL; .LS HGF and HGL FG is in the same st. line with GL; .KF is parallel to ML: Also K M has been proved parallel to FL; = two rt. angles. Wherefore, the parallelogram KFLM has been described, &c. Q.E.F. COR.-Hence a parallelogram equal to a given rectilineal figure can be applied to a given right line and in a given angle, by applying to the given right line a parallelogram equal to the first triangle. USE AND APP.-1. By this and the preceding problem we may measure the superficial contents of any rectilineal figure whatever, by first reducing it to triangles, and then making a rt. angled parallelogram equal to the sum of the triangles. We may also make a rt. angled parallelogram on a given line, and which shall be equal in area to several irregular figures. Also, if we have several figures, we may make another equal to their difference. 2. By principles already established, and by the problems for the conversion of rectilineal figures into parallelograms of equal area, we may change any rightlined figure, as ABCDE, first into a triangle, and then into a rectangle of equal area. |