Take two straight lines DE, DF, containing any angle Book VI. EDF; and upon these make DG equal to A, GE equal to B, and DH equal to C; and having joined GH, draw EF parallel a a 31. 1. to it through the point E: And because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HFb; E but DG is equal 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. b 2. 6. PROP. XIII. PROB. To find a mean proportional between two given straight lines. 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 semicircle ADC, and from the point B drawa BD * 11. 1. at right angles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angleb, and because in the right angled triangle ADC, BĎ is A drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the basec: Therefore between the two given straight lines c Cor. 8. 6. AB, BC, a mean proportional DB is found. Which was to be done. B 31. 3. м EQUAL parallelograms, which have one angle of the 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. Let AB, BC be equal parallelograms, which have the angles at B equal, and let the sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one .14. I. straight line a. The sides of the parallelograms 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 A BC, and that FE is another parallelogram, AB is to FE, 6 7. 5. as BC to FEb: But as AB to FE, so is the base DB to is the base of GB to BF; G C AB, BC about their equal angles are reciprocally proportional. 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 to BE, so is GB to BF: and as DB to BE, so is the parallelogram AB to the parallelogram FE; and as CB to BF, so is the parellelogram BC to the paral lelogram FE; therefore as AB to FE, só BC to FEd: € 9.5. Wherefore the parallelogram AB is equale to the paralle logram BC. Therefore equal parallelograms, &c. Q. E.D. Book VI. · PROP. XV. THEOR. EQUAL triangles which have one angle of the one equal to one angle of the other, have their sides 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. 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 be Let the triano, as EA 10 Aeciprocally 17.5. e 1.6. Let the triangles be placed so, that their sides, CA, AD be in one straight line; wherefore also EA and AB are in one straight line a ; and join BD. Because the triangle • 14. 1. ABC is equal to the triangle B ADE, and that ABD is another triangle: therefore as the triangle CAB, is to the triangle BAD, so is triangle AED to triangle DAB b: But as triangle CAB to triangle BAD, so is the base CA to AD«; and as triangle EAD to triangle DAB, so is the base EA to ABC: as therefore CA to AD, so is EA to ABd; wherefore the sides of the triangles ABC, 11. 5. 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, so is EA to AB; and as CA to AD, so is triangle ABC to triangle BAD ¢; and as EA to AB, so is triangle EAD to triangle BADC; therefore d 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 equale to the tri- 19. 5. angle ADE. Therefore equal triangles, &c. Q. E. D. Book VI. PROP. XVI. TIIEOR. If four straight lines be proportionals, the rectangle contained 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. Let the four straight lines AB, CD, E, F, be proportionals, viz, as AB to CD, so E to F; the rectangle con tained by AB, F is equal to the rectangle contained by * 11. 1. From the points A, C drawa AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH: Because, as AB to CD, so is E to F; and that E. is equal to CH, and 57.5. F to AG; AB isb 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 sides about equal angles reciprocally propor*14. 6. tional, are equal to one another®; therefore the parallel ogram BG is equal to the parallelogram DH: And the B "And if the rectangle contained by the straight lines, AB, I be equal to that which is contained by CD, E; these four lives are proportional, viz. AB is to CD, as E to F. The same construction heing made, because the rectangle contained by the straight lines AB, F is equal to that which is container by CD, E, and that the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogruid DH; and they are equiangular: But the sides about the Book VI. equal angles of equal parallelograms are reciprocally proportional C: Wherefore, as AB to CD, so is CH to AG; ! 4. 6. and CH is equal to E, and AG to F: as therefore AB is to CD, so is E to F. Wherefore, if four, &c. Q. E. D. PROP. XVII. THEOR. If three straight lines be proportionals, the rectangle 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. Let the three straight lines A, B, C be proportionals, viz, as A to B, so 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 a 7. 5. four straight linesa be proportionals, the A rectangle contained B by the extremes is Dequal to that which Cis contained by the means b: Therefore 16. 6. the rectangle contained by A, C is equal to that contained by B, D: But the rectangle 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 Α |