Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: the triangles ABD, ADC are similar to the whole triangle ABC, and to one another. A Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (32. 1.): therefore the triangle ABC is equiangular to.the triangle ABD, and the sides about their equal angles are proportionals (4. 6.); wherefore the triangles are similar (1. Def. 6.); in the like manner B D C it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC: and the triangles ABD, ADC, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled, &c. Q. E. D. COR. From this it is manifest, that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the segments of the base: and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side: because in the triangles BDA, ADC, BD is to DA as DA to DC (4. 6); and in the triangles ABC, DBA, BC is to BA, as BA to BD (4. 6.); and in the triangles ABC, ACD, BC is to CA as CA to CD (4. 6.). PROP. IX. PROB. FROM a given straight line to cut off any part required.* Let AB be the given straight line; it is required to cut off any part from it. A From the point A draw a straight line AC making any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it: join BC, and draw DE parallel to it: then AE is the part required to be cut off. Because ED is parallel to one of the sides of the E triangle ABC, viz. to BC, as CD is to DA, so is (2. 6.) BE to EA; and, by composition (18. 5.) CA is to AD as BA to AE: but CA is a multiple of AD; therefore (D. 5.) BA is the same multiple of AE: whatever part therefore AD is of AC, AE is the same part of AB: wherefore, from the straight line AB the part required is cut off. Which was to be done. B D C * See Note. 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. A 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 draw (31. 1.) DF EG parallels to it; and through D draw DHK parallel to AB: therefore each of the figures FH, HB is a parallelogram; wherefore DH is equal (34. 1.) to FG, and HK to GB: and because HE is parallel to KC, one of the sides of the triangle DKC, as CE to ED, so is (2. 6.) KH to HD: but KH is equal to BG, and HD to GF; therefore as CE to ED, so is BG to GF; again, because F FD is parallel to EG, one of the sides of the triangle AGE, as ED to DA, so is GF to FA; but it has been proved that 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. G B H E K C 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 (31. 1.). A Because BC is parallel to DE, a side of the triangle ADE, AB is (2. 6.) to BD, as AC to CE: but B BD is equal to AC; as therefore 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. A B C 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 having joined GH, draw EF parallel (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 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 G given straight lines A, B, C, a fourth E PROP. XIII. PROB. F 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 (11. 1.) BD at right angles to AC, and join AD, DC. D Because the angle ADC in a semicircle is a right angle (31. 3.), and be cause in the right angled triangle ADC, DB is drawn from the right angle per- A B C 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. BOOK VI. THE ELEMENTS OF EUCLID. F Complete the parallelogram FE: and because the parallelogram AB is equal to BC, and that FE is an- A other parallelogram, AB is to FE, as AB to FE, so BC to FE (7. 5.): but as 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 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. B ID 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 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 there fore CA to AD, so is EA to AB (11. C E 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. Ir 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 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 BG is contained by the straight E 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 A H B C D And if the rectangle contained by the straight lines AB, F be equal to 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. |