triangle ABC, viz. to BC, CD: DA :: BE: EA (2. 6.); and by composition (18. 5), CA : AD :: BA : AE: But CA is a multiple of AD; therefore (C. 5.) BA is the same multiple of AE, or contains AE the same number of times that AC contains AD; and therefore, whatever part AD is of AC, AE is the same of AB; wherefore, from the straight line AB the part required is cut off. PROP. X. PROB. To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have. Let AB be the straight line given to be divided, and AC the divided line, it is required to divide AB similarly to AC. F draw Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E, (Prob. 13. 1.) DF, EG, parallel to BC; and through D draw DHK, parallel to AB; therefore each of the figures FH, HB, is a parallelogram: wherefore DH is equal (27. 1.) to FG, and HK to GB: and because HE is parallel to KC, one of the sides of the triangle DKC, CE: ED: (2. 6.) KH; HD; But KH=BG, and HD = GF; therefore CE: ED :: BG: GF; Again, because FD is parallel to EG, one of the sides of the triangle AGE, ED: DA :: GF FA; But it has been proved that CE : ED :: BG: GF; therefore the given straight line AB is divided similarly A D H E K B to AC. PROP. XI. PROB. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; it is required to find a third proportional to AB, AC. Produce AB, AC to the points D, E; and make BD equal to AC; and having joined BC, through D draw DE parallel to it (Prob. 13. 1.) Because BC is parallel to DE, a side of the triangle ADE, AB : (2. 6.) BD . : AC : CE; but BDAC: therefore AB: AC:: AC: CE, Wherefore to the two given straight lines AB, AC a third proportional, CE is found. PROP. XII. PROB. To find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines; it is required to find a fourth proportional to A, B, C. Take two straight lines DE, DF, containing any angle 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 (Prob. 13. 1.) to it through the point E. And because GH is parallel to EF, one of the sides of the triangle DEF, DG: GE:: DH: HF (2. 6.); but DG=A, GE=B, and DH=C; and therefore A : B :: C: HF. Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. 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 (Prob. 6. 1.) draw BD at right angles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angle (29. 3.) and because in the right angled triangle ADC, DB is drawn from the right angle, perpendicular to the base, DB is a mean proportional between AB, BC, the seg A D B ments of the base (Cor. 8. 6.); therefore between the two given straight lines AB, BC, a mean proportional DB is found. PROP. XIV. PROB. 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 which 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. Complete the parallelogram FE; and because the parallelograms AB, BC are equal, and FE is another parallelogram, AB FE BC: FE (7. 5.): but because the parallelograms AB, FE have the same altitude, AB: FE: DB: BE (1. 6.), also, BC: FE :: GB: BF (1. 6.); therefore Wherefore, the sides of the parallelograms 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 DB: BE:: GB: BF, and DB : BE :: AB : FE, and GB: BF:: BC: EF, therefore, AB : FE :: BC: FE (11.5.): wherefore the parallelogram AB is equal (9. 5.) to the parallelogram BC. 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 (7. 1.); join BD. Because the triangle ABC is equal to the triangle ADE, and ABD is another triangle; therefore, triangle CAB: triangle BAD: triangle EAD : triangle BAD; but CAB: BAD CA: AD, and EAD: BAD: EA: AB; therefore CA: AD: EA : AB (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. B A E Having joined BD as before; because CA: AD :: EA: AB; and since CA: AD: : triangle ABC triangle BAD (1. 6.); and also EA: AB :: triangle EAD: triangle BAD (11. 5.); therefore, triangle ABC : triangle BAD triangle EAD: triangle BAD; that is, the triangles ABC, EAD have the same ratio to the triangle BAD; wherefore the triangle ABC is equal (9. 5.) to the triangle EAD. 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 E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. E From the points A, C draw (6. 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 AB CD:: E: F; and since E=CH, and F=AG, AB : CD (7. 5.) : : CH: 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 proportional, are equal to one another (14. 6.); therefore the parallelogram BG is equal to the parallelogram DH': and the parallelogram DG is contained by the straight lines AB, F; because AG is equal to F; and the parallelogram DH is contained by CD and E, because CH is equal to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. F G H And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; B C D 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 the rectangle BG is contained by AB, F, because AG is equal to F; and pathe rectangle DH, by CD, E, because CH is equal to E; therefore the rallelogram 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 AB CD:: CH: AG; but CH =E, and AG=F; therefore AB: CD:: E : F. 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 rectangle contained by the extremes is equal to that which is contained by the means (16. 6.); therefore the rectangle A.C = the rectangle B.D; but the rectangle B.D is equal to the square of B, because B= D; therefore the rectangle A.C is equal to the square of B. A B Ꭰ And if the rectangle contained by A, C be equal to the square of B; A: B:: B: 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: B:: D: C, but B=D; wherefore A: B:: B: C. 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 describe a rectilineal figure similar, and similarly situated to CDEF. Join DF, and at the points A, B in the straight line AB, make (Prob. 9. 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 (2. Cor. 25. 1.): wherefore the triangle FCD is equiangular to the triangle GAB: Again, at the points G, B in the straight line GB make (Prob. 9. 1.) the angle BGH equal to the angle DFE, and the angle GBH equal to FDE; therefore the remaining angle FED is |