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ent fides of the other; then the base that fubtends the greater Boox 1. angle of the one triangle fhall be greater than the base of the other.

Let ABC, DEF, be the two triangles, having the two fides BA, AC, equal to the two fides ED, DF, each to each, but the angle BAC greater than EDF; then the bafe BC will be greater than EF.

For, make the angle EDG equal to BAC, and DG to AC; join EG; then the bafes BC, EG, will be equal. Now, 1ft, if a 4. the right line EF fall upon EG, then EG will be greater than EF; and therefore, BC greater than EF.

b Ax. g.

2. If EF fall above EG, then F is a point within the triangle; therefore the fides DF, FE, are less than DG, GE; but DG, C 21. DF, are equal; therefore EG, or BC, is greater than EF.

d Ax. 1.

e Ax. S.

3. If EF fall below EG, join FG; then DF, DG, are equal: Therefore the angles DGF, DFG, are equal; and the whole s angle EFG greater than DFG, or DGF, and much greater than EGF; but the greater angle is fubtended by the greater fide ; 5 19. Therefore EG or BC is greater than EF. Wherefore, &c.

PRO P. XXV. THE OR.

IF
F two triangles have two fides of the one equal to two fides of
the other, each to each, and the bafe of the one greater than the
bafe of the other, the angle that the greater bafe fubtends fhall be
greater than the other.

Let the two triangles be ABC, DEF, having the sides AB, AC, equal to the two fides DE, DF, each to each, and the bafe BC greater than the base EF; then the angle BAC will be greater than the angle EDF. If not, it will be equal or lefs. If equal, the bafes BC, EF, will be equal; but they are not b. If lefs, the bafe BC will be lefs than EF; but it is not: Therefore, fince the angle BAC is neither equal nor less than EDF, it must be greater. Wherefore, &c.

I

PRO P. XXVI. T HEOR.

F two triangles have two angles of the one equal to two angles of the other, each to each, and a fide of the one equal to a fide of the other, either the fide lying between the equal angles, or fubtend

ing

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Book I. ing one of them, the remaining fides of the one triangle will be equal to the remaining fides of the other, each to each, and the remaining angle of the one equal to the remaining angle of the other.

2 3.

b 4. c Hyp.

d 4.

e Ax, 1.

f 16.

& IG.

b Hyp.

Let the two triangles be ABC, DEF, having the two angles ABC, ACB, of the one, equal to DEF, DFE, of the other, each to each.

1. Let the fide BC be equal to EF, viz. the fides lying between the equal angles; then the fides BA, AC, will be equal to the fides ED, DF, each to each; and the angles BAC, EDF, equal. For, if the fide AB be not equal to DE, let one of them, as AB, be the greater; from which cut off GB equal to DE, join GC; then, fince GB, BC, are equal to DE, EF, and the angles GBC, DEF, equal, the bases GC, DF, are equal; and the angles GCB, DFE, equal'; but the angle DFE is equal to ACB: Therefore GCB is equal to ACB, a part to the whole; which is impoffible: Therefore GB is not equal to DE, nor is any fide but AB equal to DE: Therefore AB, BC, are equal to DE, EF; the angle ABC, to DEF; and the bafe AC to DF d.

a

2. Let the fides AB, DE, which fubtend the equal angles, be equal; if any of the fides, as BC, be not equal to EF, let BC be the greater; cut off BH equal to EFa; join AH; then, because AB, BH, are equal to DE, EF, and the angle ABH to DEF, the base AH equal to DF, and the angle AHB to DFE; but the angle ACB is equal to DFE: Therefore the angle AHB is equal to ACB, and likewife greaterf; which is impoffible. Wherefore, &c.

IF

PRO P. XXVII. THE OR.

a right line fall upon two right lines, making the alternate angles equal, these right lines will be parallel.

Let the right line EF fall upon the two right lines AB, CD, making the angles AEF, EFD, equal, the right lines AB, CD, will not meet one another, whether produced towards B, D, or A, C. Let them be produced; and, if poffible, meet in the point G; then EGF is a triangle; the outward angle AEF is greater than EGF, or EFG: but AEF, EFG, are equal", and likewife greater; which is impoffible: Therefore AB, CD, will not meet, if produced toward B, D. For the fame reafon they will not meet, if produced toward A, C: Wherefore AB, CD, are parallel.

PRO P. XXVIII. T HEOR.

Fa right line fall upon two right lines, making the outward

inward angles on the fame fide equal to two right angles; these two right lines fhall be parallel.

Let the right line EF fall upon the two right lines AB, CD, making the outward angle EGB equal to the inward and oppofite GHD; or the inward angles BGH, GHD, together, equal to two right angles; then AB, CD, will be parallel.

Book I.

d 27.

For, becaufe the angles EGB, GHD, are equal, AGH is a hyp. equal to EGB; and therefore equal to GHD: therefore AB b 15. is parallel to CD d. Again, because the angles BGH, GHD, are c Ax. 1. equal to two right ones '; but AGH, BGH, are equal to two right angles; therefore AGH, BGH, are equal to BGH, GHD c. c 13, Take the common angle BGH from both, the remainders AGH, GHD, are equal f; but these are alternate angles: Therefore f Ax. 3′′ AB is parallel to CDd. Wherefore, &c.

7

PRO P. XXIX. THE O R.

IF Fa right line fall upon two parallel lines, the alternate angles will be equal; the outward angle equal to the inward and oppofite, on the fame fide; and the two inward angles on the same fide equal to two right angles.

For, let EF fall upon the two parallèl lines AB, CD, the alternate angles AGH, GHD, will be equal; the outward angle EGB equal to the inward GHD; and the two inward angles BGH, GHD, equal to two right angles.

a Cor. 17.,

For, if the angle AGH is not equal to GHD, let one of them be greater, as AGH; then the right lines AB, CD, produced toward B, D, will meet one another in fome point; but they are parallel; therefore cannot meet: Therefore AGH is b Def. 35. not greater than GHD. For the fame reason it is not lefs; therefore it is equal. But EGB is equal to AGH; therefore e 15. EGB is equal to GHD d. Add BGH to both; then EGB, d Ax. I, BGH, are equal to BGH, GHD; but EGB, BGH, are equal e Ax. 2. to two right angles f: Therefore BGH, GHD, are equal to two f 13. right angles. Wherefore, &c.

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Book I.

2 29°

b Ax. I.

C 27.

a 23.

b Conft. € 27.

R

PRO P. XXX. THE OR.

IGHT lines parallel to one and the fame right line, are pa rallel to one another.

Let AB, CD, be two right lines, each parallel to EF; AB will be parallel to CD.

Let GK fall upon them; then, because GK falls upon the parallels AB, EF, the angles AGH, GHF, are equal. Again, because GK falls upon the parallels EF, CD, the outward angle GHF is equal to the inward and oppofite GKD; therefore the angles AGK, GKD, are each equal to GHF; therefore equal to one another : Therefore AB is parallel to CD c. Wherefore, &c.

PRO P. XXXI. PRO B.

O araw a right line through a given point parallel to a given

T right line.

It is required, through the point A, to draw a right line parallel to the right line BC.

Affume any point D, in BC; join AD; and make the angle DAE equal to the angle ADC; join EA, and produce it to F; then the alternate angles EAD, ADC, are equal to one another: Therefore EF, BC, are parallel. Wherefore, &c.

PRO P. XXXII. T HEOR.

IF one fide of a triangle be produced, the outward angle is equal to both the inward oppofite angles; and the three inward angles are equal to two right angles.

Let ABC be a triangle, CD a fide produced; the outward angle ACD is equal to the inward and oppofite angles ABC, BAC; and the three angles ACB, ABC, and BAC, are together equal to two right angles. Through C draw CE parallel to AB; then the angle BAC is equal to ACE; and the angle ECD, to ABC; therefore the whole angle ACD is equal to the two angles ABC, BAC. Add the angle ACB to both; then the two angles ACD, ACB, are equal to the three angles ABC,

ACB, BAC; that is, equal to two right angles. Where- Book I. fore, &c.

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1. Hence all the three angles of any one triangle are b Ax. I. equal to all the three angles of any other triangle, either fepa- 13. rately or taken together.

2. If two angles of one triangle be equal to two angles of another triangle, either feparately or together, the remaining angle of the one is equal to the remaining angle of the other.

3. If one angle of a triangle be a right one, the other two angles are together equal to a right angle.

4. If the angle included by the equal fides of an ifofceles triangle be a right one, each of the other angles will be half a right one.

5. Any angle in an equilateral triangle is one third of two right angles, or two thirds of one right angle.

6. If one angle of a triangle be equal to the other two, that angle is a right one; for, if the fide is produced, the adjacent angle is equal to the other two; therefore each of them are right angles.

7. All the inward angles of any right lined figure make twice as many right angles, abating four, as the figure has fides. For any right lined figure can be divided into triangles, the inward angles of each equal to two right angles, and all the triangles together equal to the number of fides of the figure, abating two Therefore all the inward angles will be equal to twice the number of fides, abating four.

8. All the outward angles of any right lined figure are equal to four right angles. For all the outward and inward angles together are equal to double the number of fides; but the inward angles are equal to double the number of fides, abating four: Therefore the outward are equal to four right angles.

PRO P. XXXIII. THEOR.

F two right lines join two equal and parallel right lines toward the fame part, thefe lines will be equal and parallel.

Let AB, CD, be two equal and parallel right lines; join AC, BD; then will the right lines AC, BD, be equal and parallel.

For, because AB, CD, are parallel, and BC falls upon them, the angle ABC is equal to BCD; but, because AB is equal to a 29. CD, and BC common; and the angle ABC equal to BCD, the bafe AC is equal to BD, and the angle ACB equal to CBD b; b 4.

c

but

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