* 7. 5. * 2. 6. equal to BG, and HD to GF; therefore, as CE to ED, so is BG to GF: again, because FD is parallel to GE, one of the sides of the triangle AGE, as ED-to DA, so is GF to FA: therefore, as has been proved, CE is to ED, as BG to GF; and as ED to DA, so GF to FA. Therefore the given straight line AB is divided similarly to AC. Which was to be done. PROP. XI. PROB. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines; it is required to find a third proportional to AB, AC. Let AB, AC be placed so as to contain any angle; produce AB until BD, the part produced, is equal to AC; join BC, and through D, 31. 1. draw* DE parallel to BC, meeting AC B produced in E: then CE shall be a third proportional to AB and AC. * 2. 6. * 7.5. 3. 1. A C of the triangle ADE, AB is to BD, as AC to CE: but BD is equal to AC; therefore* as AB 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. 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: join GH, and HF; but DG is equal to A, GE to B, and DH to C; * 2. 6. therefore, as A is to B, so is C to HF: wherefore to 7. 5. the three given straight lines, A, B, C a fourth 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 AC describe the semicircle ADC, and from the point B draw BD at right angles to AC: then BD shall be a mean proportional between AB and BC. Join AD, DC; then the angle A 11. 1. ADC, being in a semicircle is a right angle; and be- * 31. 3. cause in the right-angled triangle ADC, DB is drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between AB, BC the segments of the base:* therefore between the two Cor. 8. given straight lines AB, BC, a mean proportional DB is found. Which was to be done. 6. 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 of 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 * 14. 1. line; wherefore also a FB, BG are in one straight line:* complete the parallelogram FE; and because the parallelogram AB is equal to BC, * 7. 5. * 1. 6. and that FE is another parallelo- * D F G F is the base GB to BF; therefore, as DB to BE, so is 11. 5. GB to BF; wherefore the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. * 1. 6. 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 • Because the angles FBD, FBE are together equal to two 13. 1. right angles,* and that the angle DBF is equal to the angle EBG, by hypothesis; therefore the two angles FBE, EBG are 14. 1. together equal to two right angles, and consequently FB, BG are in the same straight line. 11.5. to * 9. 5. to BE, so is the parallelogram AB to the parallelogram 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 shall be reciprocally proportional; that is, CA shall be to AD, as EA to AB. Let the triangles be placed so that their sides CA, AD be in one straight line; where forea also EA and AB are in one * 14. 1. * 7. 5. straight line; and join BD. Because the triangle ABC is equal to the triangle ADE, and that ABD is another triangle; therefore as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle DAB:* but as the triangle CAB to the triangle BAD, so is the base CA to the base AD;* and as the * 1. 6. triangle EAD to the triangle DAB, so is the base EA to the base AB; therefore as CA to AD, so is EA to * 1. 6. AB; wherefore the sides of the triangles ABC, ADE * 11. 5. about the equal angles are reciprocally proportional. Next, let the sides of the triangles ABC, ADE about See the note to the last 'Proposition. • Hyp. 1. 6. 1. 6. 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 BAD;* and as EA to AB, so is the triangle EAD to the triangle BAD;* therefore as the triangle BAC to the triangle BAD, so is the 11. 5. triangle EAD to the 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. Therefore, equal triangles, &c. Q. E, D. ⚫ 9.5. 11. 1. 3. 1. 7.5. 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, shall be equal to the rectangle contained by CD, E. From the points A, C draw AG, CH at right angles to AB, CD; and make* AG equal to F, and CH * 31. 1. 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; therefore* AB is to CD, as CH to AG: that is, the sides of the parallelograms BG, DH about the equal angles are reciprocally pro portional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another; therefore the parallelogram BG is equal to the parallelogram DH: and the parallelogram 14. 6. |