See N. Book VI. PRO P. VII. THEOR. angle of the other, and the sides about two other angles proportionals; then if each of the remaining angles be either less, or not less than a right angle; or if one of them be a right angle: the triangles shall be equiangular, and have those angles equal about which the sides 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 sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remaining angles at C, F be less than a right angle. 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 straight line AB make the angle ABG equal to the A a. 23. 1. angle DEF, and because the D G B CE b. 32. I. angle AGB is equal b to the remaining angle DFE. therefore the 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 Hypothesis, is AB to BC; therefore as AB d. 11. s. to BC, so is AB to BG d; and because AB has the same ratio to each c. 9. 5. of the lines BC, BG; BC is equal to BG, and therefore the angle f. 5. 1. BGC is equal to the angle BCG f. but the angle BCG is, by Hy pothesis, less than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater 6. 13•• than a right angle 8. 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 Hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, that c. 4. 6. is, they are equal. and the angle at A is equal to the angle at D; Book VI. 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 tri angie DEF. The same construction being made it may be proved in A D like manner that BC is equal to BG, and the angle at Cěqual to the angle BGC. but G B the angle at C is not less than с E a right angle; therefore the angle BGC is not lefs than a right angle. wherefore two angles of the triangle BGC are together not less than two right angles; which is impossiblen; and therefore the triangle ABC may be proved to be e- h. 14. 1. quiangular to the triangle DEF, as in the first cafe. Laitly, 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 equiangular, 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 B cafe, that BG is equal to BC. C D but the angle BCG is a right angle, therefore f the angle BGC f. s. so is also a right angle; whence two of the angles of the tri Е Е angle BGC are together not less C than two right angles ; which is impoffible , therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if two triangles, &c. Q. E. D. А A PROF. Book VI. Sec N. PROP. VIII. THEOR. 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 perpendicuiar 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 to the A a. 32. 1. remaining angle BAD. there fore the triangle ABC is equian- the sides about their equal angles D C c. 1. Def. 6. triangles are similar c. in the like manner it may be demonstrated that the triangle ADC is similar to the triangle ABC. Also the triangles ABD, ADC are similar to one another. Because the right angle BDA is equal to the right angle ADC, and also 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 similar to the triangle ADC. 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 DCb; and in the triangles ABC, DBA, BC is to BA, as BA to BDb; and in the triangles ABC, ACD, BC is to CA, as CA to CD b. PROP Book VI. PROP. IX. PROB. FROM "ROM a given straight line to cut off any part required. See N. 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 with А E D required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, a. 2. 6. fo is . BE to EA; and, by composition, CA b. 18. S. is to AD, as BA to AE. but CA is a multiple of AD, therefore o BA is the same multiple of B C c. D. s. 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. TO divide a given straight line fimilarly to a given di vided 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 A fimilarly to AC. Let AC be divided in the points D, E; and let AB, AC be placed so as to F D contain any angle, and join BC, and H! through the points D, E draw * DF, G EG parallels to it; and through D draw B DHK parallel to AB. therefore each of K the figures FH, HB is a parallelogram; wherefore a. 31. 1. Book VI. wherefore DH is equal o to FG, and HK to GB. and because HE is parallel to KC one of the sides of the b. 34. 1. triangle DKC, as CE to ED, so is © KH t. 2. 6. to HD. but KH is equal to BG, and HD to GF; therefore as CE to ED, so F D is BG to GF. again, because FD is pa HI E E К. C ΤΟ find a third proportional to two given straight lines. B Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; it is required A 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, 2. 31. 1. through D draw DE parallel to it". Because BC is parallel to DE a side of the b. 2. 6. triangle ADE, AB is b to BD, as AC to CE, but BD is equal to AC, as therefore AB to D ΤΟ O 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 |