equal to the angle DCE, and the angle ACB to the angle, DEC, and consequently (32. 1.) the angle BAC equal to the angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals ; and those are the homologous sides which are opposite to the equal angles. Let the triangle DCE be placed, so that its side CE may be conti- F D B С E And because AC is parallel to FE, one of the sides of the triangle FBE, BA : AF :: BC : CE (2. 6.) : but AF is equal to CD; therefore (7. 5.), BA : CD :: BC : CE ; and alternately, BA : BC : DG :: CE (16. 5.): Again, because CD is parallel to BF, BC : CE :: FD: DE (2. 6.); but FD is equal to AC ; therefore BC : CE :: AC : DE : and alternately, BC : CĂ :: CE : ED. Therefore, because it has been proved that AB : BC :: DC : CE ; and BC: CA :: CE : ED, ex æquali, BA : AC : : CD : DE. Therefore the sides, &c. Q. E. D. PROP. V. THEOR. if the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides. Let the triangles ABC, DEF have their sides proportionals, so that, AB is to BC, as DE to EF ; and BC to CA, as EF to FD; and consequently, ex æquali, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC being equal to the angle DEF, and BCA to EFD, and also BAC to EDF. por aolo C G At the points E, F, in the straight line EF, make (23. 1.) the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the remaining angle BAC is equal to E F the remaining angle EGF (32.1.), and the triangle ABC is therefore equiangular to the triangle GEF ; and consequently they B have their sides opposite to the equal angles proportionals (4. 6.). Wherefore, AB: BC :: GÈ: EF ; but by supposition, DE : EF :: GE : EF. Therefore (11. 5.) DE and GE have the same ratio to EF, ind consequently are equal (9. 5.). For the same reason, DF is equal to FG : And because, in the triangles DEF, GEF, DE is equal to EG, and EF common, and also the base DF equal to the base GF ; therefore the angle DEF is equal (8. 1.) to the angle GEF, and the other angles to the other angles, which are subtended by the equal sides (4. 1.). Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF : and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC ; therefore the angle ABC is equal to the angle DEF: For the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D. PROP. VI. THEOR. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the hoinologous sides. Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF. and ACB to DFE. At the points D, F, in Α. the straight line DF, make (23. 1.) the angle FDG equal to either of the angles D BAC, EDF; and the angle G DFG equal to the angle ACB ; wherefore the remaining angle at B is equal to the remaining one at G B E (32. 1.), and consequently the triangle ABC is equiangular to the triangle DGF ; and therefore BA: AC :: GD (4. 6.) : DF. But by hypothesis, ED: DF :: GD: (11. 5.) DF; wherefore ED is equal (9. 5.) to DG : and DF is common to the iwo triangles EDF, GDF; therefore the two sides ED, DF are equal to the two sides GD, DF ; but the angle EDF is also equal to the angle GDF ; wherefore the base : EF is equal to the base FG (4. 1.), and the triangle EDF to the tri angle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides : Therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E : But the angle DFG is equal to the angle ACB ; therefore the angle ACB is equal the angle DFE, and the angle BAC is equal to the angle EDF (Hyp.) ; wherefore also the remaining angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equian: gular to the triangle DEF. Wherefore, if two triangles, &c. Q. E, D PROP. VII. THEOR. 1 two triangles have one angle of the one equal to one 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, 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 AFC is equiangular to the triangle DEF, that is, the angle APC is equal to the angle DEF, and the remaining angle at C to the remaining angle at F. For, if the angles AEC, 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 angle (23. 1.) A DEF : and because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF; the remaining angle AGB is equal 6 СЕ wherefore (4. 6.), AB : BG :: DE: EF ; but, DE : F:: AB : BC, therefore, AB : BC :: AB : BG (11. 5.) ; and because AB has the same ratio to each of the lines BC, BG ; BC is equal (9.5.) to BG, and therefore the angle BGC is equal to the angle BCG (6. 1.): But the angle BCG is, by hypothesis, less than a right angle ; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle (13. 1.). 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 is, 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 less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF. The same construction being made, it may be proved, in like manner, that BC is equal to BG, and the angle at C equal to Α. the angle BGC : But the an D gle at C is not less less than a right angle; therefore the angle BGC is not less than a G right angle : Wherefore, two B C angles of the triangle BGC are F together not less than two right angles, which is impossible (17. 1.); and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. E PROP. VIII. THEOR. 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 the angle at B 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 proportion B D c als (4. 6.) ; wherefore the triangles are similar (def. 1. 6.). In the like manner, it may be demonstrated, that the triangle ADC is equian: gular and similar to the triangle ABC : And the triangles ABD, ADC, being each equiangular and similar to ABC, are equiangular and similar to one another. 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. For in the triangles BDA, ADC, BD : DA :: DA : DC (4. 6.) ; and in the triangles ABC, BDA, BC : BA :: BA : BD (4.6.) ; and in the triangles ABC, ACD, BC : CA :: CA : CD (4. 6.). PROP. IX, PROB. From a given straight line to cut off any part required, that is, a pare which shall be contained in it a given number of times. Let AB be the given straight line ; it is required to cut off from AB, a part which shall be contained in it a given number of times. From the point A draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC such that it shall contain AD, as oft as AB is to А. contain the part, which is to be cut off from it i join BC, and draw DE parallel to it: then AE is the part required to be cut off. D Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, CD : DA :: DE : EA (2. 6.); and by composition (18.5.), CA : AD :: BA : AE : But CA is a multiple of AD ; therefore (C. 5.) BA is the same multiple of AE, or contains AE the same num B ber of times that AC contains AD; and there С fore, whatever part AD is of. AC, AE is the same of AB ; wherefore, from the straight line AB the part required is cut off. Which was to be done. 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 (31. 1.) DF, EG, parallel to BC ; and through D draw DHK parallel to AB ; therefore each of the figures FH, HB, is a parallelo |