Take two straight lines DE, DF, containing any angle EDF: and upon these make DG equal to A, GE D equal to B, and DH equal to C; and A having joined GH, draw EF parallel (31. 1.) to it through the point E: B and because GH is parallel to EF, C С one of the sides of the triangle DEF, DG is to GE, as DH to HF (2. 6.); 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 F given straight lines A, B, C, a fourth E proportional HF is found. Which was to be done. 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 AO describe the semicircle ADC, and from the point B draw D (1l. 1.) BD at right angles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angle (31. 3.), and because in the right angled triangle ADC, DB is drawn from the right angle per- A B С pendicular to the base, DB is a mean proportional between AB, BC, the segments of the base (Cor. 8. 6.): therefore between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done. PROP. XIV. THEOR. 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 straight line (14. 1.): 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 BC, and that FE is an. A F other parallelogram, AB is to FE, as BC to FE (7. 5.): but as AB to FE, SO is the base DB to BE (1. 6.); and as E BC to FE, so is the base GB to BF: D B therefore as DB to BE, so is GB to BF (11. 5.). Wherefore the sides of the parallelograms AB, BC about their equal angles are reciprocally propor G C tional, 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, só is the parallelogram AB to the parallelogram FE; and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FE (9. 5.): wherefore the parallelogram AB is equal (9. 5.) to the parallelogram BC. Therefore, equal parallelograms, &c. Q. E. D. 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 is, CA is to AD, as EA to AB. 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 (14. 1.) and join BD. Because the triangle ABC is equal to the triangle ADE, and that ABD is another B D triangle, therefore as the triangle CAB is to the triangle BAD, so is the triangle EAD to triangle DAB (7.5.): but as triangle CAB to triangle BAD, SO is the base CA to AD (1. 6.): and as triangle EAD to triangle DAB, so is the base EA to AB (1. 6.): as therefore CA to AD, so is EA to AB (11. C 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 so is EA to AB; and as CA to AD, so is triangle BAC to triangle BAD (1. 6.); and as EA to AB, so is triangle EAD to triangle BAD (1. 6.); therefore (11. 5.) 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 (9. 5.) to the triangle ADE. Therefore, equal triangles, &c. Q. E. D. PROP. XVI. THEOR. 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 is E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. From the points A, C draw (11. 1.) 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 F to AG; AB is (7. 5.) to CD, as CH to AG: therefore the sides of the parallelograms B DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14. 6.); therefore the parallelogram BG is equal to the parallelogram DH, and the parallelogram BG is contained by the straight E lines AB, F, because AG is equal H to F; and the parallelogram DH is F contained by CD and E, because CH is equal to E; therefore the G rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. And if the rectangle contained by the straight lines AB, F. be equal to A в с D 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 contained by the straight lines AB, F is equal to that which is contained 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 parallelogram DH, and they are equiangular : but the sides about the equal angles of equal parallelograms are reciprocally proportional (14. 6.); wherefore, as AB to CD, so is CH to AG; and CH is equal to E, and AG to F: as therefore AB is to CD, so 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 (7. 5.) to B, as D to C; but if four straight lines be proportionals, the A rectangle contained by the В extremes is equal to that D which is contained by the C means(16. 6.): therefore the D rectangle contained by A, C is equal to that contained by B, D. But the rectangle contained by B, D is B 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 (16. 6.); therefore A is to B, as D to C; but B is equal to D; wherefore as A to B, so B to C. Therefore, if three straight lines, &c. Q. E. D. PROP. XVIII. PROB. Upon a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.* Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB to deseribe a rectilineal figure similar and similarly situated to CDEF. Join DF, and at the points A, B, in the straight line AB, make (23. 1.) the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the remaining angle AGB (32. 1.); wherefore the triangle FCD is equi * See Note. angular to the triangle GAB: H. again at the points G, B, in G the straight line GB, make, F E (23. 1.) the angle BGH equal to the angle DFE, and the angle GBH equal to FDE: K therefore the remaining angle FED is equal to the re- A B С D maining angle GHB; and the triangle FDE equiangular to the triangle GBH: then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole CFE: for the same reason, the angle ABH is equal to the angle CDE; also the angle at A is equal to the angle at C, and the angle GHB to FED; therefore the rectilineal figure ABHG is equiangular to CDEF: but likewise these figures have their sides about the equal angles proportionals ; because the triangles GAB, FCD being equiangular, BA is (4. 6.) to AG, as DC to CF; and because AG is to GB, as CF to FD; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to FE; therefore, ex æquali (22. 5.), AG is to GH, as CF to FE: in the same manner it may be proved that AB is to BH, as CD to DE: and GH is to HB, as FE to ED (4. 6.). Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another (1. def. 6.). Next, let it be required to describe upon a given straight line AB, a rectilineal figure similar and similarly situated to the rectilineal figure CDKEF. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar and similarly situated to the quadrilateral figure CDEF, by the former case; and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK; therefore the re remaining angle at K is equal to the remaining angle at L: and because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED, and BHL is equal to DEK; wherefore the whole angle GHL is equal to the whole angle FEK : for the same reason the angle ABL is equal to the angle CDK; therefore the five sided figures AGHLB, CFEKD are equiangular; and because the figures AGHB, CFED are similar, GH is to HB, as FE to ED; and as HB to HL, so is ED to EK (4. 6.); therefore, ex æquali (22. 5.), GH is to HL, as FE to EK: for the same reason, AB is to BL, as CD to DK: and BL, is to LH, as (4. 6.) DK to KE; because the triangles BLH, DKE are equiangular; therefore, because the five sided figures AGHLB, CFEKD are equiangular, and have their sides about the equal angles proportionals, they are similar to one another; and in the same manner a rectilineal figure of six or more sides may be described upon a given straight line similar to one given, and so on. Which was to be done. |