* 31.1. A Let AB, AC be placed so as to contain any angle: produce AB, AC, to the points D, E; and make BD equal to ACį. join BC, and through D, draw DE parallel to it *: CE shall be a third proportional to AB and AC. Because BC is parallel to DE, a side of the BH АС triangle ADE, AB is to BD, as AC to CE: but BD is equal to AC; therefore as AB is to † AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Which was to be done. . 2. 6. + 7. 5. D PROPOSITION XII. • 31.1. ProB. - To find a fourth proportional to three given straight lines. Take two straight lines, DE, DF, containing any angle D A B с JI * 2. 6. E 7.5. PROPOSITION XIII. lines. Place AB, BC in a straight line, and upon B C tween AB and BC. Join AD, DC: and because the angle ADC in a semi . 11. 1. eirele is a right angle, and because in the right-angled tri- • 31. 3. angle ADC, BD is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base * Therefore, between the two * Cor. 8. 6. given straight lines AB, BC, a mean proportional DB is found. Which was to be done. PROPOSITION XIV. E Thor.-Equal parallelograms which hare 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: the sides of the parallelograms AB, BC about the equal angles, shall be reciprocally proportional ; that is, DB shall be to BE, as GB to BF. Let the sides DB, BE be placed in the same straight line; wherefore also $ FB, BG F are in one straight line * ; complete the * 14. I. D parallelogram FE: and because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is to FE, as BC to FE*: but as AB to FE, so is the base DB to BE*, and . 7. 5. as BC to FE, so is the base GB to BF; therefore, as DB to BE, so is GB to BF* Wherefore, the sides of the parallelo- • 11.5. grams AB, BC about their equal angles are reciprocally proportional. Next, let the sides about the equal angles be reciprocally proportional, viz, as DB to BE, so GB to BF: the parallelogram AB shall be equal to the parallelogram BC. Because, as DB to BE, so is GB to BF; and as DB to BE +, + 1.0. so is the parallelogram AB to the parallelogram FE ; and as GB to BF, so is the parallelogram BC to the parallelogram B . 1. 6. Because the angle DBF is equal † to the angle GBF, add to each the + Hyp. angle FBE; therefore the two angles DBF, FBE are together equal t to the + 2 Ax. two angles FBE, EBG: but DBF, FBE are together equal † to two right † 13. 1. angles; therefore FBE, EBG are together equal † to two right angles : + 1 Ax. therefore FB, BG are in the same straight line. * 14. 1. L • 11.5. 9.5. FE; therefore as AB to FE, so BC to FE: therefore the parallelogram AB is equal * to the parallelogram BC. Therefore, equal parallelograms, &c. Q. E. D. PROPOSITION 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. • 14. 1. 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 shall be reciprocally proportional ; that is, CA shall be to AD, as EA to AB. Let the triangles be placed so that their B sides CA, AD be in one straight line; where+ See N. last fore + also EA and AB are in one straight Prop. line *; and join BD. Because the triangle с ABC is equal to the triangle ADE, and that ABD is another triangle, therefore as the triangle CAB, is to the • 7.5. triangle BAD *, so is the triangle AED to the triangle DAB: but as the triangle CAB to the triangle BAD, so is the base • 1. 6. CA to AD, and as the triangle EAD to the triangle DAB, • 1.6. so is the base EA to AB* ; therefore as CA to AD *, so is EA • 11.5. to AB: wherefore the sides of the triangle ABC, ADE, about the equal angles, are reciprocally proportional. Next, 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 shall be equal to the triangle ADE. Join BD as before : then because, as CA to AD, so is EA to AB; and as CA to AD, so is the triangle ABC to the triangle . 1.6. BAD *; and as EA to AB, so is the triangle EAD to the tri. 1. 6. angle BAD*; therefore as the triangle BAC to the triangle • 11.5. BAD, so is the triangle EAD to the triangle BAD ; that is, the triangles BAC, EAD have the same ratio to the triangle • 9. 5. BAD: wherefore the triangle ABC is equal * to the triangle ADE. Therefore, equal triangles, &c. Q. E. D. PROPOSITION 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, shall be equal to the rectangle contained by CD, E. From the points A, C, draw * AG, CH at right angles to AB, * 11. 1. CD, and make + AG equal to F, and ch equal to E; and com- † 3. ). plete + the parallelograms BG, DH. Be + 31, 1. cause, as AB to CD, so is E to F; and that E H F E is equal to CH, and F to AG, AB is * to • 7.5. G CD as CH to AG: therefore the sides of the parallelograms BG, DH, about the equal angles, are reciprocally proportional ; but A parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another*; therefore the parallelogram BG is equal to the parallelo- • 14. 6. gram DH: but the parallelogram BG 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. And if the rectangle contained by the straight lines AB, F, be equal to that which is contained by CD, E, these four lines shall be proportional, viz. AB shall be 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 t to the parallelogram DH; and they are + 1 Ax. equiangular : but the sides about the equal angles of equal parallelograms are reciprocally proportional* : wherefore, as AB to CD, so is CH to AG : but ch is equal to E, and AG to F; therefore as AB is to CD +, so is E to F. Wherefore, if + 75. four, &c. Q. E. D. 14. 6. PROPOSITION XVII. 7.5. A • 16. 6. 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 te 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, shall be 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 * 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 D the means *; therefore the rectangle contained by A, C, is equal to that contained by B, D: but the rectangle B contained by B, D, is the square because B is equal to D; therefore the rectangle contained by À, C, is equal to the square of B. And if the rectangle contained by A, C, be equal to the square of B, A shall be 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 * : 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. A of B, . 16. 6. Q. E. D. See N. 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 rectilincal figure of four sides ; it is required, upon the given |