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 A Let A, and D, be so placed that two sides of one square may be in the same straight lines with two sides of the other, each to each; and produce the other sides of A and D until they meet. The resulting fig, will be a square, composed of the two squares A and D, together with two equal rectangles, E, E. Then since A and E are parallelograms of the same altitude, base of A: base of E. A: E Also E: D:: base of A: base of E; .. A: E :: E : D. The same construction being made with B and C, as in the annexed fig., it may be shewn in the same manner that B: F :: F: C. Now let a, b, c, d, e, f represent straight lines having the same ratio to each other as the areas A, B, C, D, E, F. Then, since A: B:: a: b, CD cd, and A: B :: C : D, E d = E D .. a b c d; and therefore the rectangle contained by a and d = the rectangle contained by b and c. Similarly, aeed, and b:ƒ::ƒ: c; .. rectangle a, square of e, and rectangle b, c= square of f; .. square of e = square off, and .. e=ƒ (42 Cor. 3). But E: Fe: f, .. E = F. Now E is the rectangle contained by a side of A and a side of D; and F is the rectangle contained by a side of B and a side of C; and rectangle E = rectangle F; therefore (74 Cor. 1), side of A side of B :: side of C: side of D. COR. 1. Hence also the converse is easily shewn, viz. that, if four straight lines are proportionals, the squares described upon them are proportionals. COR. 2. In (76) it was proved that the areas of similar triangles are to one another as the squares of corre may be done here without inconvenience, because we are not much concerned with the magnitude of the sides until we arrive at the last stage of the proof. sponding, or homologous', sides. It may now be shewn that the areas are proportional to the squares of the altitudes also of the triangles, or to the squares of any corresponding lines within the triangles. For (see fig. in 76) CD: cd BC: bc = AB: ab, square of CD: square of cd = square of AB : square of ab; and .. triangle ABC : triangle abc square of CD: square cd. = 80. PROP. XIII. If there be any number of magnitudes, which, taken two and two, have a certain fixed ratio to each other, the sum of the first terms of the several pairs of magnitudes shall be to the sum of their second terms in that same ratio. Let A, B, C, D be four magnitudes such that A: B :: E: F, and C: D:: E: F, then also A+C: B+ D :: E : F. For suppose a, b, c, d, e, f to represent six straight lines having the same ratio to each other that A, B, C, D, E, F, have. Then since a:be:f, the rectangle a, f= the rectangle b, e (74). Similarly, the rectangle c, f= the rectangle d, e. .. rectangle a, ƒ + rectangle c, f= rectangle b, e + rectangle d, e. H Now, constructing each of these rectangles, as in the annexed figs. by making AB= a, BC=c, CD=f; also EF = b, FG=d, GH =e; it is obvious that the rectangle a, ƒ+ rectangle c, f= rectangle contained by AB + BC, and CD, that is, rectangle a +c, f. B E F G Also rectangle b, e + rectangle d, e = rectangle contained by EF + FG, and GH, that is, rectangle b+d, e. .. rectangle a + c, f= rectangle b + d, e, But a+c: b+ d :: A + B : B + D, and ef E: F, by supposition, ..A+C: B+D: E: F. |