first and fourth is equal to the rectangle contained by the second and third.* A Let A, B, C, D be four straight lines, 'proportionals', that is, A is to В as C to D. The rectangle contained by A and D shall be equal to the rectangle contained by B and C. Draw the straight line EF equal to A; produce it to G making FG equal to B; through Fdraw FH, FK at right angles to EF, making FH-C, and FK =D; through E and G draw EM, LGN parallel to HK; and through H, and K draw HL, and MKN parallel to EFG. Then EFKM, FGNK, and FHLG are all rectangular parallelograms, B as will easily appear. Now since EFKM and FGNK are parallelograms between the same parallels, EFKM : FGNK :: EF : FG (73), that is :: A : B. But A: B= C: D, since A, B, C, D are proportionals, .. EFKM : FGNK :: C : D. Again, FHLG : FGNK :: FH : FK, that is :: C: D, :. EFKM : FGNK :: FHLG : FGNK, which signifies that EFKM is the same multiple, part, or parts of FGNK that FHLG is of the same magnituđe FGNK, .. it is plain that EFKM=FHLG. But EFKM is the rectangle contained by EF, FK, that is, A and D; also FHLG is the rectangle contained by FG, FH, that is, B and C; .. the rectangle contained by A and D is equal to the rectangle contained by B and C. This is sometimes expressed by saying, 'if four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means'. COR. 1. Conversely if A, B, C, D be any four straight lines, such that the rectangle contained by A and D is equal to the rectangle contained by B and C, then A, B, C, D are proportionals. COR. 2. If A, B, C, D be proportionals, since the rectangle B, C=the rectangle A, D, it follows, from Cor. 1, that B, A, D, C are proportionals, that is, B AD C. This change of position in the different members of a proportion is called 'invertendo', or 'by inversion'. COR. 3. Since the rectangle contained by C and B is equal to the rectangle contained by B and C, if A, B, C, D are proportionals, it follows that the rectangle A, D=the rectangle C, B, and .. A, C, B, D are proportionals, that is, A: C:: B: D. This is called 'alternando', or 'alternately'. COR. 4. Since the measure of the ratio of one area to another is simply the number of times the one contains or is contained in the other, this ratio may always be represented by the ratio of one line to another. Hence the two preceding Corollaries hold also for areas as well as lines, that is, if A and B be two areas, and C and D two lines, or if A, B, C, D be four areas, such that A B C D, then inversely B: A: D: C. Also if A, B, C, D be four areas proportionals, such that A : B :: C: D, then, alternately, A: C :: B : D. 75. PROP. IX. The areas of parallelograms, or triangles, on the same base, are proportional to their altitudes'. D I I (1) Let ABCD, ABEF be two parallelograms, on the same base AB; from any point G in AB draw GIH at right angles to AB, meeting CD in H, and EF in I. Then GH is the altitude of the parallelo gram ABCD, and GI the altitude of ABEF. And ABCD shall be to ABEF as GH is to GI. Let such a lineal unit of measurement be taken as will exactly divide both GH and GI; and divide GH and GI into as many equal parts as they contain this unit. Through the several points of division draw lines parallel to AB, dividing ABCD into as many parallelograms as the unit is contained in GH, and ABEF into as many as the unit is contained in GI. These smaller parallelograms are obviously all equal to one another; and therefore the area ABCD will be to the area ABEF in the same ratio as the number of them in the former area is to the number in the latter, that is, as the number of units in GH is to the number in GI, that is, as GH is to GI. (2) Again, join BD, BF; then ABD, ABF will represent any two triangles on the same base AB. From D and F draw DK, FL perpendiculars to AB; then DK, FL are the 'altitudes' of the triangles ABD, ABF. Also DK=GH, and FL = GI (35 Cor.) Now ABD is half of the parallelogram ABCD; and ABF is half of ABEF; and the halves of two magnitudes must obviously have the same ratio to one another which the whole magnitudes have; .. the triangle ABD : the triangle ABF:: GH: GI, that is, :: DK: FL; or the triangles are proportional to their 'altitudes'. COR. It follows also that parallelograms, or triangles, upon equal bases are proportional to their altitudes. 76. PROP. X. The areas of similar triangles are proportional to the squares of any two corresponding sides*, that is, sides opposite to equal angles+. Let ABC, abc be similar triangles, in which A = 2α, Sometimes called 'homologous sides'. +Euclid's enunciation of this is: Similar triangles are to one another in the duplicate ratio of their homologous sides'. LB=Lb, ¿C= L c; then AB, ab being any two corresponding, or homologous, sides, the triangle ABC shall be to the triangle abc as the square of AB is to the square of ab. Upon AB describe the square AEFB, and upon ab the square aefb. From C draw CD perpendicular to AB, and from c draw cd perpendicular to ab. Through C draw GCH parallel to AB, and through c draw gch parallel to ab. Then, since a triangle is always equal to half the parallelogram upon the same base and between the same parallels, triangle ABC: triangle abc :: paral". AGHB: paralTM.aghb. Now paralm. AGHB: square of AB :: AG: AE, i. e. CD: AB, (73), and paral. aghb: square of abag: ae, i. e. :: cd: ab; but CD AB=cd: ab (71, Cor. 3, and 74, Cor. 3), since ABC, abc, are similar triangles, ·. paralTM. AGHB : square of AB :: paralTM, aghb and, alternately, : square paral". AGHB: paral". aghb :: square of AB .. triangle ABC: triangle abc :: square of AB of ab; : square of ab, : square of ab. [This is one of the most important Theorems in Geometry.] 77. PROP. XI. To find a fourth proportional to three given straight lines, that is, a fourth line such that the four lines shall be proportionals. Let A, B, C be the three given straight lines; it is required to find another X, such that A: B :: C : X. Draw any indefinite straight line DEF, in which take DE equal to A, and EF equal to B. From D draw A B C DGH making any angle with DF, in which take DG equal to C. Join EG, and through F draw FH parallel to EG, meeting the line DGH in H. 0 Then since DFH is a triangle, and EG is parallel to FH, DE: EF :: DG : GH (70). But DE=A, EF=B, DG=C; . A: B: C: GH, that is, GH = X, the straight line required. COR. By the same method a third proportional may be found to two given straight lines, that is, a third line C such, that A : B :: B : C. The only difference is, that DG is taken equal to EF F and equal to B. Then DE: EF:: DG: GH, that is, A: B :: B : GH, .. GH = C. E A G H 78. DEF. Four-sided figures are similar, when they have their angles equal, each to each, and the sides forming equal angles proportionals. Hence all squares are similar figures, since the angles of any one are equal to the angles of any other, each to each, and the sides about equal angles (being equal) are proportionals. But neither two rectangles, nor two parallelograms with angles equal each to each, are necessarily similar. In addition to the equality of angles, the sides about the equal angles must be proportionals; and although, in the case of triangles, it follows as a consequence of the equality of angles, that the sides about equal angles are proportionals (71), yet it is not so with any other rectilinear figures, except squares, as may easily be shewn. For instance, if a part be cut off from a parallelogram by a straight line parallel to one of the sides, the new parallelogram will have its angles equal to those of the original one, each to each; but it is obvious that the sides about equal angles in each are not proportionals; and therefore the parallelograms are not similar. 79. PROP. XII. If the squares described upon four straight lines be proportionals, the straight lines themselves are proportionals; and conversely. Let A, B, C, D* represent four squares, proportionals, that is, A: B :: C : D. * In this proposition a single letter is used to designate a square or a rectangle, contrary to rule, merely to avoid unnecessary writing. This |