Book I.

COR. Hence BD, DC, are not equal to BA, AC, each to each. Wherefore, if in any case it is thought neceffary to prove that part of Prop. VII. when the one point falls within the triangle, it is evident from this.

T

PRO P. XXII. PRO B.

O make a triangle, whofe fides are equal to three given right lines, if any two of them, however taken, are greater than the third.

Let A, B, C, be the three given right lines, any two of which are greater than the third. Take any right line bounded at D, but not bounded at E, from which cut off DF equal to A, FG equal to B, and make GH equal to C; then, with the center F, and diftance DF, defcribe the circle DKL; with the center G, and distance GH, describe the circle KLH; from the point K, where the circles cut each other, draw the right lines. Def. 15. FK, KG; then FD is equal to FK ; but FD is equal to A therefore FK is equal to A. For the fame reafon GK is equal to C, and FG is equal to B: Therefore the three fides FK, FG, GK, of the triangle FKG, are equal to the three given right lines, A, B, C. Wherefore there is conftitute, &c.

8.

A

PRO P. XXIII. PRO B.

Ta given point, in any right line, to make an angle equal to a given right lined angle.

Let A be the given point in the right line AB; it is required to make an angle equal to the right lined angle DCE.

Affume any points D, E, in the right lines CD, CE, and join DE. At the point A, in the line AB, make a triangle AFG, whofe fides are equal to the three right lines CD, CE, DEa; then, because the two fides GA, AF, are equal to the two fides 'CE, CD, each to each, and the bafes GF, ED, equal, the angles GAF, ECD, are equal. Wherefore there is conftitute, &c.

I

PRO P. XXIV. T HE OR.

F two triangles have two fides of the one cqual to two fides of the other, each to cach, and the angle contained by the two fides of the one greater than the angle contained by the correfpond

ent

ent sides of the other; then the base that fubtends the greater Book L angle of the one triangle shall 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, ift, if a 4. the right line EF fall upon EG, then EG will be greater than EF b; and therefore, BC greater than EF.

.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, DF, are equal; therefore EG, or BC, is greater than EF.

d Ax. I.

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 f s angle EFG greater than DFG, or GF, and much greater than EGF; but the greater angle is fubtended by the greater fide; 8 3Hi Therefore EG or BC is greater than EF. Wherefore, &c.

PRO P. XXV. T H E O R.

I1
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
base of the other, the angle that the greater bafe fubtends ball be
greater than the other.

Let the two triangles be ABC, DEF, having the fides AB, AC, equal to the two fides DE, DF, each to each, and the bafe BC greater than the bafe 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. 4 'b Hyp If lefs, the base BC will be lefs than EF; but it is not Cat Therefore, fince the angle BAC is neither equal nor lefs than EDF, it must be greater. Wherefore, &c.

PRO P. XXVI. THE OR.

b

IF
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

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.

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 DE2, 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 base AC to DF d.

2. Let the fides AB, DE, which subtend 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 EF; join AH; then, because AB, BH, are equal to DE, EF, and the angle ABH to DEF, the bafe 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.

Fa 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 reason they will not meet, if produced toward A, C: Wherefore AB, CD, are parallel.

PRO P.

Book I.

PRO P. XXVIII. THE OR.

F a right line fall upon two right lines, making the outward angle equal to the inward and oppofite, on the fame fide, or the 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.

d 27.

For, because the angles EGB, GHD, are equal, AGH is a hyp. equal to EGB b; and therefore equal to GHD: therefore AB b 15. is parallel to CD4. Again, because the angles BGH, GHD, are c Ax. I. equal to two right ones ; but AGH, BGH, are equal to two right angles; therefore AGH, BGH, are equal to BGH, GHD c. e 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 CD4. Wherefore, &c.

PRO P. XXIX. THE OR.

C

IF a 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 fame fide equal to two right angles.

For, let EF fall upon the two parallel 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.

b Def. 35.

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 not greater than GHD. For the fame reafon it is not less; therefore it is equal. But EGB is equal to AGH; therefore c 15. EGB is equal to GHD d. Add BGH to both; then EGB, d Ax, 1. 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.

PROP.

BOOK I.

a 29.

b Ax. I.

C 27.

a 23.

b Conft.

C 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 parasiel to CD.

Let GK fall upon them; then, becaufe 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. 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 pa. rallel 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.

2 29.

I

PRO P. XXXII. T HEOR.

F 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 thetwo angles ACD, ACB, are equal to the three angles ABC,

ACB,