PROPOSITION IX. PROBLEM. From a given right line to cut off any part required. Let AB be a given right line; it is required to cut from AB any part required.* Let a third part be required, and from it draw any right line AC, containing any angle with AB, and take any point D in AC and make DE, EC, equal to AD, join BC, and through D draw DF parallel to BC. And because FD is drawn parallel to one of the sides BC of the triangle ABC; therefore, proportionally as CD is to Da so is BF to FA. But CD is double of DA; therefore also BF is double of FA; hence BA is triple of AF. F B A D E a 4. 6. Therefore, from the given right line AB the third part required has been cut off, viz. AF. Q. E. F. PROPOSITION X. PROBLEM. To divide a given right line into parts similarly to a given divided right line. Let AB be the given divided right line, also ac cut in the points D, E, and place them so that they may contain any angle, and join CB, and through D, E, draw DF, EG, parallel to BC, and through D draw DHK 31. 1. parallel to AB. F G A H D Therefore each of the figures FH, HB, is a parallelogram; hence DH is equal to FG, also HK to GB. And because HE is drawn parallel to one of the sides KC of the triangle DKC; proportionally as CE is to ED so is KH to HD; but кH is equal to BG; also HD to GF; therefore as CE is to ED so is BG to GF. Again, because FD is drawn parallel to one of the sides EG of the triangle AGE; hence proportionally it is as ED to DA SO is GF to FA. But it has been de B K C *Precisely in the same manner may any part whatever be cut off, since AF in all cases is the same part of AB which AD is of ac. . 31. 1. b 2.6. monstrated as CE is to ED so is BG to GF; therefore as CE is to ED so is BG to GF, also as ED to DA so is GF to FA. Therefore a given right line, &c. Deductions. Q. E. D. 1. To describe a square which shall have a given ratio to a given rectilineal figure. 2. To divide a right line into three parts which shall be in harmonical progression. 3. The base, the vertical angle, and the ratio of the two sides of a triangle being given to construct it. PROPOSITION XI. PROBLEM. Two right lines being given, to find a third proportional. Let AB, AC, be the given right lines, and place them so that they contain any angle; it is required to find a third proportional. For produce AB, AC, to the points D, E, and place BE equal to AC, also join BC, and through D draw DE parallel to it.a D A A E Therefore BC is drawn parallel to one of the sides DE of the triangle ADE, proportionally it is as AB to b BE SO is AC to CD. But BE is equal to AC, therefore it is as AB to AC so is AC to BE. Therefore two lines AB, AC, being given, a third proportional BE has been found. Q. E. F. Deduction. To determine the locus of the vertices of all the triangles, which can be described on a given base, so that each of them shall have its two sides in a given ratio. PROPOSITION XII. PROBLEM. Three given right lines being given to find a fourth proportional to them. Let A, B, C, be three given right lines; it is required to find a fourth proportional to them. Place the two right lines DE, DF, containing any angle EDF, and make DG equal to A, also GE equal to B, and DH equal to c; GH being joined, draw through E, EF parallel to it.a D A E H FABC • 31. 1. And because GH is drawn parallel to one of the sides EF of the triangle DEF, therefore as DG is to GE so is DH to HF. But DG is equal to A, also GE to B, and DH ↳ 2. 6. to c; therefore as A is to в so is c to HF. Therefore three right lines being given A, B, C, a fourth proportional HF has been found. Q. E. F. Deductions. 1. Divide a given right line into two parts, so that the rectangle contained by them may be equal to a given rectangle. 2. From a given point to draw a right line to cut a given circle, so that the distances of the two intersections from the given point, shall be to each other in a given ratio. PROPOSITION XIII.* PROBLEM. Two right lines being given to find a mean proportional. Let AB, BC, be two given right lines; it is required to find a mean proportional to AB, BC. Put them in a right line, and upon Ac describe the semi-circle ADC, and draw from the point, B, BD at right angles to the right line Ac‚a and join AD, DC. And because the angle ADC in a semi-circle is a right angle. And because in the right angled triangle ADC, the perpendicular DB is drawn from the right angle to the base; DB is A * 11. 1. D b 31. 3. B с a mean proportional between the segments of the base. © Cor. 3. 6. Therefore the two right lines AB, BC, being given, a mean proportional BD has been found. Q. E. F. * This is in effect the same as the last proposition of the second book. a 14. 1. b 7.5. c 11. 5. d 9, 5. PROPOSITION XIV. THEOREM. If equal parallelograms have one angle of the one equal to one angle of the other, the sides about the equal angles are reciprocally proportional; and if parallelograms have one angle of the one equal to one angle of the other, and the sides about the equal angles reciprocally proportional; these parallelograms shall be equal to one another. a F A Let AB, BC, be equal parallelograms having the angles at в equal, and place DB, BE, in a direct line, therefore FB, BG, are in a direct line; then the sides about the equal angles of the parallelograms AB, BC, are reciprocally proportional, that is, as DB to BE so is GB to BF, For complete the parallelogram FE. And because the parallelogram AB is equal to the parallelogram BC, and FE E is some other parallelogram; therefore as AB to FE so is BC to FE. But as AB to FE so is DB to BE, also as BC to FE so is GB to BF; and therefore as DB to BE so is GB to BF.C Therefore the sides of the parallelograms AB, BC, are reciprocally proportional. C B G But also let the sides about the equal angles be reciprocally proportional, viz. as DB to BE So is GB to BF; then the parallelogram AB is equal to the parallelogram BC. For because it is as DB to BE SO is GB to BF, but as DB to BE so is the parallelogram AB to the parallelogram FE, also as GB to BF so is the parallelogram BC to the parallelogram FE; hence also as AB to FE so is BC to FE, therefore the parallelogram AB is equal to the parallelogram BC. Therefore if equal parallelograms, &c. Q. E. D. PROPOSITION XV. THEOREM. If equal triangles have one angle of the one equal to one angle of the other, the sides about the equal angles are reciprocally proportional; and if triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles reciprocally proportional, these triangles are equal. Let ABC, ADE, be equal triangles having one angle, BAC equal to an angle DAE; then the sides of the triangles ABC, ADE, are reciprocally proportional, that is, as CA to AD So is EA to AB. For place ac in a direct line with AD, therefore EA is in a direct line with And join BD. A B.a And because the triangle ABC is equal to the triangle ADE, but ABD is another triangle; therefore as the triangle CAB is to the triangle BAD so is the triangle ADE to the triangle BAD." But as CAB to BAD So is CA to AD, also as EAD is to B BAD so is EA to AB; therefore the sides of the triangles ABC, ADE, are reciprocally proportional. a 14. 1. b 7.5. Dc 1. 6. C E Next let the sides of the triangles ABC, For BD being joined, because it is as CA to AD so is EA to AB, but as CA to AD so is the triangle BAC to the triangle BAD, also as EA to AB so is the triangle EAD to the triangle BAD; therefore as the triangle ABC to BAD SO is the triangle EAD to BAD; hence each of them ABC, ADE, has the same ratio to BAD; therefore the triangle ABC is equal to the triangle EAD. Therefore if equal triangles, &c. Q. E. D. PROPOSITION XVI.* THEOREM. If four right lines be proportional, the rectangle contained under the extremes is equal to the rectangle contained under the means; and if the rectangle contained under the extremes be equal to the rectangle contained under the means, the four right lines will be proportional. Let AB, CD, E, F, be four proportional right lines, viz. as AB to CD so is E to F; then the rectangle contained under AB, F is equal to the rectangle contained under CD, E. For draw AG, CH, at right angles to the lines AB, CD,a from the points A, C, and make AG equal to F, also CH equal to E, and complete the parallelograms BG, DH. *Algebraically if A: B: C: D, then AD = BC. For since AD BC by multiplying by BD. a a 11. 1. |