Take two straight lines DE, DF containing any angle EDF; and Book VI. upon these make DG equal to A, GE equal to B, and DH D A equal to C; and having joined GH, draw EF parallel a to B a. 31.1. it through the point E. and С becautė GH is parallel to EF one of the sides of the tri G H angle DEF, DG is to GE, as DH to HFb. but DG is equal E F to A, GE to B, and DH to C; therefore as A is to B, so is C to HF. Wherefore to the three given straight lines A, B, C a fourth proportional HF is found. Which was to be done. PRO P. XIII. PROB. ΤΟ "O find a mean proportional between two given straight lines. D Let AB, BC be the two given straight lines; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the femicircle ADC, and from the point B draw * BD at right angles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angles, and be b. 31. 3. cause in the right angled triangle ADC, DB is drawn from the right A B C angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the bafe. therfore between the two given straight lines AB, BC, a meaac. Cor.6. proportional DB is found. Which was to be done. THEOR. Book VI. PROP. XIV. one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. c. 1. 6. Let AB, BC be equal parallelograms which have the angles at B equal, and let the sides DB, BE be placed in the fame straight line ; à. 14. 1. wherefore also FB, BG are in one straight line', the sides of the pa rallelograms AB, BC about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF. Complete the parallelogram FE; and because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is A F $. 2. . to FE, as BC to FEb. but as AB GB to BF; therefore as DB to d. 11.5. BE, fo.is GB to BF4. Wherefore the fides of the parallelo G But let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF; the parallelogram AB is equal to the parallelogram BC. Because as DB is to BE, so GB to BF; and as DB to BE, so is the parallelogram AB to the parallelogram FE; and as GB to BF, fo is parallelogram BC to parallelogram FE; therefore as AB to FE, ¢. Ø. S. fo BC to FEd. wherefore the parallelogram AB is equal to the pa rallelogram BC. Therefore cqual parallelograms, &c. Q. E. D. PROP Book Ví. PROP. XV. THEOR. EQUAL triangles which have one angle of the one equal to one angle of the other, have their fides about the equal angles reciprocally proportional: and triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another. b. 7.5. C. I. 6. Let ABC, ADE be equal triangles which have the angle BAC equal to the angle DAE; the sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to AB. Let the triangles be placed so that their fides CA, AD be in one straight line; wherefore also EA and AB are in one straight linea; 2. sa. 1. and join BD. Because the triangle ABC is equal to the triangle ADE, and that ABD is another triangle; therefore as the triangle CAB B D is to the triangle BAD fo is triangle EAD to triangle DAB b. but as triangle CAB to triangle BAD, so is A the base CA to AD“; and as triangle E AD to triangle DAB, so is the base EA to ABC; as there С fore CA to AD, so is EA to ABC. . 11.5 wherefore the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional. But let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC is equal to the triangle ADE. Having joined BD as before, because as CA to AD, fo is EA to AB; and as CA to AD, fo is triangle BAC to triangle BAD"; and as EA to AB, so triangle EAD to triangle BAD'; therefore è as triangle BAC to triangle BAD, so is triangle EAD to triangle BAD; that is, the triangles BAC, EAD have the same ratio to the triangle BAD. wherefore the triangle ABC is equal to the triangle ADE. 4.9. .. Therefore equal triangles, &c. Q. E. D. PROP. XVI. THEOR. tained by the extremes is equal to the rectangle contained by the means: and if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. a. II. I. Let the four straight lines AB, CD, E, F be proportionals, viz. as AB to CD, fo E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. From the points A, C draw a AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the pi rallelograms BG, DH. because as AB to CD, so is E to F; and that .B.9.5. E is equal to CH, and F to AG; AB is b to CD, as CH to AG, therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their lives about equal angles reciprocally proportional, are equal 6. 14. 6. to one another; therefore the parallelogram BC is equal to the ri rallelogram DH, and the pa- H F G A B C D equal to thai whică is contained by CD and E. And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; these four lines are proportionals, viz. AB is to CD, as E to F. The same construction being made, because the rectangle confuíned by the straight lines AB, F is equal to that which is contained ky €, E, and that the rectangle BG is contained by AB, F, becole AG is equal to F; and the rectangle DH by CD, E, because CH is equal to É; therefore the parallelogram EG is equal to the maalle ogram DH; and they arc equiangular. but the sides about the the equal angles of equal parallelograms are reciprocally proportio- Book VI. nalc. wherefore as AB to CD, fo is CH to AG; and CH is equal to m E, and AG to F. as therefore AB is to CD, fo E to F. Wherefore -: 14 6. if four, &c. Q. E. D. PROP. XVII. THEOR. IF .contained by the extremes is equal to the square of the mean: and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. b. 16. 6. Let the three straight lines A, B, C be proportionals, viz. as A to B, fo B to C; the rectangle contained by A, C is equal to the square of B. Take D equal to B; and because as A to B, so B to C, and that B is equal to D; A is a to B, as D to C. but if four straight 2. 7. 5. lines be proportionals, the rectangle contained A by the extremes is equal B to that which is con- D tained by the means b. C therefore the rectangle contained by A, C is e D 0 qual to that contained by B, D. but the rect A. B angle contained by B, D is the square of B; because B is equal to D. therefore the rectangle contained by A, C is equal to the square of B. And if the rectangle contained by A, C be equal to the square of B; A is to B, as B to C. The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D; therefore the rectangle contained by A, C is equal to that contained by B, D. but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals b. therefore A is to B, as D to C; but B is equal to D; L 3 wherefore |