Book VI. See N. 2. 23. I. b. 32. I. c. 4. 6. d. II. S. e. 9. 5. f. 5. I. 8. 13. 1. IF F two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportionals; then if each of the remaining angles be either lefs, or not less than a right angle; or if one of them be a right angle. the triangles fhall be equiangular, and have those angles equal about which the fides are proportionals. Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the fides about two other angles ABC, DEF proportionals, fo that AB is to BC, as DE to EF; and, in the first cafe, let each of the remaining angles at C, F be less than a right angie. The triangle ABC is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C, to the remaining angle at F. For if the angles ABC, DEF be not equal, one of them is greater than the other; let ABC be the greater, and at the point B in the ftraight line AB make the an A triangle ABG is equiangular to the triangle DEF; wherefore as AB is to BG, fo is DE to EF; but as DE to EF, fo, by Hypothefis, is AB to BC; therefore as AB to BC, fo is AB to BG; and because AB has the fame ratio to each of the lines BC, BG ; BC is equal to BG, and therefore the angle BGC is equal to the angle BCGf. but the angle BCG is, by Hypothefis, lefs than a right angle; therefore alfo the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle . But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle. but, by the Hypothefis, it is lefs than a right angle; which is ab furd. Therefore the angles ABC, DEF are not unequal, that is, Book VI. they are 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 lefs than a right angle. the triangle ABC is also in this cafe equiangular to the triangle DEF. The fame conftruction being made, it may be proved in like manner that BC is equal to BG, and the angle at C equal to the angle BGC. but the angle at C is not lefs than a right 44 CE angle; therefore the angle B BGC is not lefs than a right angle. wherefore two angles of the triangle BGC are together not lefs than two right angles; which is impoffible ; and therefore the triangle ABC may be proved to h. 17. 1. be equiangular to the triangle DEF, as in the firft cafe. Lastly, Let one of the angles at C, F, viz. the angle at C be a right angle; in this cafe likewife the triangle ABC is equiangular to the triangle DEF. For if they be not equiangular, make at the point B of the ftraight line AB the angle ABG equal to the angle DEF; then it may be proved, as in the first cafe, that BG is equal to BC. but the angle BCG is a right angle, therefore f the angle BGC is alfo a right angle; whence two of the angles of the triangle BGC are together not lefs than twop right angles; which is impoffi ble b. therefore the triangle ABC A is equiangular to the triangle DEF. Wherefore if two triangles, &c. Q. E. D. Book VI. Zee N. 2. 32. I. PROP. VIII. THE OR. IN a right angled triangle, if a perpendicular be drawn from the right angle to the bafe; the triangles on each fide 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 bafe BC. the triangles ABD, ADC are fimilar 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 re maining angle ACB is equal to the remaining angle BAD1. therefore the triangle ABC is equiangular to the triangle ABD, and the fides about their equal angles are proportionals b, wherefore the c. 1. Def. 6. triangles are fimilar c. in the like b. 4. 6. B A D C manner it may be demonftrated that the triangle ADC is fimilar to the triangle ABC. Alfo the triangles ABD, ADC are fimilar to one another. Because the right angle BDA is equal to the right angle ADC. and alfo the angle BAD to the angle at C, as has been proved; the remaining angle at B is equal to the remaining angle DAC. therefore the triangle ABD is equiangular and fimilar to the triangle ADC. Therefore in a right angled, &c. Q. E. D. COR. From this it is manifeft that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the fegments of the bafe. and alfo that each of the fides is a mean proportional between the bafe, and its fegment adjacent to that fide. becaufe 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 CD, PROP. IX. PROB. Book VI. 'ROM a given ftraight line to cut off any part re- See N. FRO quired. Let AB be the given ftraight line; it is required to cut off any part from it. From the point A draw a straight line AC making any angle with AB; and in AC take any point D, and take AC which is the fame 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 fame part of AB that AD is of AC; that is, AE E is the part required to be cut off. aB D C a. 2. 6. Because ED is parallel to one of the fides of the triangle ABC, viz. to BC, as CD is to DA, fo is BE to EA; and, by composition b, CA is to AD, as BA to AE. but CA is a multiple of AD, therefore BA is the fame multiple of AE. whatever part therefore AD is of AC, AE is the fame part of AB. wherefore from the straight line AB the part required is cut off. Which was to be done. O divide a given ftraight line fimilarly to a given divided straight line, that is, into parts that shall have the fame ratios to one another which the parts of the divided given ftraight line have. Let AB be the straight line given to be divided, and AC the divided line; it is required to divide AB fimilarly to AC. Let AC be divided in the points D, E; and let AB, AC be placed fo as to contain any angle, and join BC, and F through the points D, E draw DF, G EG parallels to it; and through D draw DHK parallel to AB. therefore each of B the figures FH, HB is a parallelogram. A H K c. D. 5. b. 34. I. c. 2. 6. A Book VI. wherefore DH is equal to FG, and HK to GB. and becaufe HE is parallel to KC one of the fides of the triangle DKC, as CE to ED, fo is KH to HD. but KH is equal to BG, and HD to GF; therefore as CE to ED, fo is BG to GF. again, because FD is parallel to EG one of the fides of the triangle AGE, as ED to DA, fo is GF to FA. but it has been proved that CE is to ED, as BG to GF; as therefore CE a. 31. 1. b. 2. 6. F K to ED, fo is BG to GF; and as ED to DA, fo GF to FA. therefore the given ftraight line AB is divided fimilarly 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 ftraight lines, and let them be placed fo 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, thro' D draw DE parallel to it ". B A C D Because BC is parallel to DE a fide of the triangle ADE, AB is b to BD, as AC to CE. but BD is equal to AC, as therefore AB to AC, fo 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. E "O find a fourth proportional to three given ftraight lines. Let A, B, C be the three given flraight lines; it is required to find a fourth proportional to A, B, C. |