< B = <b, and < C = c; it is required to shew that ab: AB: ac: AC, ab: AB: bc: BC, and bc BC:: ac: AC. : With centre A and radius ab describe a circle cutting AB in D, and another circle with the same centre and radius ac cutting AC in E, and join ED. Then in the two triangles ADE, abc, the two sides AD, AE, are equal to the two sides ab, a c, each to each, and ‹DAE +bac,.. the triangles are equal in all respects (24), and.. ADE=4abc. But abc = LABC, .. L LADE = ABC, and .. DE is parallel to BC (34). Hence by (70 Cor.) AD : AB :: AE : AC; but AD = ab, and AE= = ac, . ab AB :: ac: AC. In the same way, by making B the centre of the circles, it may be shewn, that ab AB bc: BC; and by making C the centre, that bc: BC ac: AC. COR. 1. Conversely, if two triangles have an angle of one equal to an angle of the other, and the sides forming the equal angles proportionals, the triangles will be similar. COR. 2. Hence, also, if a triangle be cut off from a larger triangle by a line parallel to one of the sides, the two triangles will be similar, and have the sides about equal angles proportionals. COR. 3. If CD be drawn perpendicular to AB, and cd to ab, it will also easily appear, that CD: cd :: AC: ac, or :: AB: ab, or :: BC: bc. 72. PROP. VI. If a right-angled triangle be divided into two other right-angled triangles by a straight line drawn from the vertex of the right angle perpendicular to the opposite side, each of these two triangles shall be similar to the whole triangle and to one another. Let ABC be a right-angled triangle having <BAC the right angle. From A draw AD perpendicular to BC; then the triangles ABD, ADC shall be similar to the triangle ABC, and to each other. Because BAC = a right angle =<ADB, and B is common to the two triangles ADB, ABC; and since the three angles of every triangle are together equal to two right angles, (37), the remaining 4 BAD of the one triangle is equal to the remaining ACB of the other, that is, the triangles ADB, ABC, are equiangular, and.. similar. B D In the same way it may be shewn, that the triangles ADC, ABC, are equiangular, and .. similar. Hence also the triangles ADB, ADC are equiangular and similar. COR. 1. Since in similar triangles the sides forming equal angles are proportionals (71); and since the triangles ADB, ADC are similar, .. BD : AD :: AD : DC. Again, since ABD and ABC are similar triangles, BD AB :: AB : BC. : Also, since ADC and ABC are similar, CD: AC AC: BC. [DEF. When of four magnitudes which are proportionals the second and third are the same, this latter magnitude is said to be a mean proportional between the other two. Thus, in this Cor. AD is a mean proportional between BD and DC. Also AB is a mean proportional between BD and BC; and AC is a mean proportional between BC and CD.] COR. 2. Hence to find a mean proportional between two given straight lines, AB, BC, place AB, BC so as to form one straight line AC. Upon AC describe a semicircle; through B draw BD at right angles to AC meeting the circumference in D; and join AD, CD. Then, since ▲ ADC is a right A B C angle, and DB is perpendicular to AC, by Cor. 1, AB: BD:: BD: BC; :. BD is a mean proportional between AB and BC. 73. PROP. VII. The areas of parallelograms or triangles between the same parallels are proportional to their bases. D C (1) Let ABCD, BEFC be two parallelograms between the same parallels ABE, DCF, and upon the bases AB, BE, respectively. B E Let such a lineal unit of measurement be taken as will exactly divide both AB, and BE; and divide AB and BE A into as many equal parts as they contain the unit (68). Through the several points of division draw lines parallel to AD or BC, dividing ABCD into as many parallelograms as the unit is contained in AB, and BEFC into as many as the unit is contained in BE. Then since parallelograms upon equal bases and between the same parallels are equal to one another (41 Cor. 1), the smaller parallelograms which make up ABCD, and BEFC, are all equal. Therefore the ratio of ABCD to BEFC will be the ratio of the sum of these equal parallelograms in the one to the sum of them in the other, that is, as the number of them in the one to the number in the other, (since they are all equal) or as the number of units in AB is to the number of units in BE, that is, as AB is to BE. (2) Again, since a parallelogram is double of the triangle upon the same base and between the same parallels (40), and the halves of two magnitudes will plainly bear the same ratio to each other that the whole magnitudes do,.. joining BD, EC, the triangle ABD, which is half of the parallelogram ABCD, will have the same ratio to the triangle BEC, which is half of BEFC, that AB has to BE. COR. It follows also, that the areas of triangles or parallelograms of equal altitudes, however situated, are proportional to their bases; the altitude being the perpendicular let fall from the vertex of one of the angles upon the opposite side considered as the base. 74. PROP. VIII. If four straight lines taken in order be proportionals', the rectangle contained by the 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. |