fore the angles ABC, DEF are not unequal, that is, they are Book VI. equal : and the angle at A is equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F: therefore the triangle ABC is equiangular to the triangle DEF. Next, Let each of the angles at C, F, be not less than a right angle: the triangle ABC is also in this case equiangular to the triangle DEF. The same construction being A made, it may be proved in like D manner that BC is equal to BG, and the angle at C equal to the G angle BGC: but the angle at C is not less than a right angle; therefore the angle BGC is not B с Е F less than a right angle: wherefore two angles of the triangle BGC are together not less than two right angles, which is impossible b; and therefore the triangle ABC may be proved to be h 17. 1. equiangular to the triangle DEF, as in the first case. Lastly, Let one of the angles at C, F, viz. the angle at C, be a right angle; in this case likewise the triangle ABC is equiangular to the triangle DEF. For, if they be not equiangu A lar, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC: B D but the angle BCG is a right angle, therefore i the angle BCC А i3 4 is also a right angle; whence two of the angles of the triangle BGC are together not less E F than two right angles, which is impossibleh: therefore the tri. B с angle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D. с Book VI. PROP. VIII. THEOR. See N IN a right angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another. 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. 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 А a 32. 1 to the remaining angle BAD a: therefore the triangle ABC is equi- the sides about their equal angles b 4. 6. are proportionalsb; wherefore the cl.def.co triangles are similare: in the like B D C manner 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 b; and in the triangles ABC, DBA, BC is to BA, as BA to BD b; and in the triangles ABC, ACD, BC is to CA, as CA to CDb. Book VI. PROP. IX. PROB. FROM a given straight line to cut off any part See N. required. Let AB be the given straight line; it is required to cut off any part from it. From the point A draw a straight line AC making any angle Because ED is parallel to one of the sides E D b 18. 5. tiple of AD; thereforec BA is the same mul cD. 5. tiple of AE: whatever part therefore AD is of AC, AE is the same part of AB: where B с fore, from the straight line AB the part required is cut off. Which was to be done. a 2.6. 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. 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 a DF, EG parallels to it; and through D a 31. 1 . draw DHK parallel to AB: therefore each of the figures FH, HB, is a parallelogram ; wherefore DH is equal b to FG, and b 34. 1. Book VỊ. HK to GB: and because HE is paral А lel to KC, one of the sides of the trian. c 2. 6. gle DKC, as CE to ED, so is c KH to HD: but KH is equal to BG, and HD D F H E B K 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 A C a 31. 1. through D draw DE parallel to it a. B but BD is equal to AC; as therefore AB to AC, E 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 ; Book VI. and upon these make DG equal D * to A, GE equal to B, and DH equal to C; and having joined GH, draw EF parallel a to it B a 31. 1. through the point E: and be с cause GH is parallel to EF, one of the sides of the triangle G DEF, DG is to GE, as DH to н b 2.6 HFb; but DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. E F Wherefore to the three given straight lines A, B, C a fourth proportional HF is found. Which was to be done. 1 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 a BD at right an D a 11. 1. gles to AC, and join AD, DC. Because the angle ADC in a b 31. 3. semicircle is a right angle b, and because in the right angled triangle ADC, DB is drawn from the right angle perpendicular to A B C the base, DB is a mean proportional between AB, BC, the segments of the base c: therefore be-c Cor.8. tween the two given straight lines AB, BC a mean proportional & DB is found. Which was to be done. Y |