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BOOK I.

Let AB be the given ftraight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A, in the given straight line AB, that fhall be equal to the given rectilineal angle DCE.

Take in CD, CE, any points

a 22. 1. D, E, and join DE; and make a the triangle AFG, the fides of which thall be equal to the three ftraight lines CD, DE, CE, fo that CD be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG,

A

D

B/

A

each to each, and the bafe DE to the base FG; the angle DCE b 8. 1. is equal b to the angle FAG: Therefore, at the given point A, in the given ftraight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.

a 23. 1. b 3. 1.

€ 4. I.

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PROP. XXIV. THEOR.

F two triangles have two fides of the one equal to two fides of the other, each to each, but the angle contained by the two fides of one of them, greater than the angle contained by the two fides. equal to them, of the other; the bafe of that which has the greater angle, fhall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is alfo greater than the bafe EF.

Of the two fides DE, DF, let DE be the fide which is not greater than the other, and at the point D, in the ftraight line DE, make a the angle EDG equal to the angle BAC; and make DG equal b to AC or DF, and join EG, GF.

Becaufe AB is equal to DE; and AC to DG, the two fides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is alfo equal to the angle EDG; therefore the bafe BC is equal to the bafe EG; and becaufe DG is equal to DF, d 5.1. the angle DFG is equal d to the angle DGF; but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and that

the

the greater e fide is oppofite to the greater angle; the fide EG is therefore greater than the fide EF; but EG is equal to BC; and therefore alfo BC is greater than EF. Therefore, if two triangles, &c. Q E. D.

IF

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PROP. XXV. THEOR.

[F two triangles have two fides of the one equal to two fides of the other, each to each, but the bafe of the one greater than the bafe of the other; the angle alfo contained by the fides of that which has the greater bafe, fhall be greater than the angle contained by the fides equal to them, of the other.

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the bafe CB is greater than the bafe EF; the angle BAC is likewife greater than the angle EDF.

D

For, if it be not greater, it must either be equal to it, or lefs; but the angle BAC is not equal to the angle EDF, becaufe then the bafe BC would be equal a to EF; but it is not; therefore the angle BAC is not equal to the angle DEF: neither is it lefs; because then the base BC would be lefs b than the bafe EF; but it is not; therefore

the angle BAC is not lefs B

than the angle EDF: and

CE

Book I.

e 19. 1.

a 4. I.

b 24. I.

F

it was fhewn that it is not equal to it; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D.

PROP.

Book I.

I

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F two triangles have two angles of the one equal to two angles of the other, each to each; and one fide equal to one fide, viz. either the fides adjacent to the equal angles, or the fides oppofite to equal, angles in each: then fhall the other fides be equal, each to each; and alfo the third angle of the one to the third angle of the other.

A

D

Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD; also one fide equal to one fide; and first let those fides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF; the other fides fhall be equal, each to each, viz. AB to DE, and AC to DF; and thet hird angle BAC to the third angle EDF.

G

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CE

F

For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, and BC to EF, the two fides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is also equal to 4.1. the angle DEF; therefore the bafe GC is equal a to the base DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal fides are oppofite; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothefis, equal to the angle BCA; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is impoffible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; the base therefore AC is equal a to the base DF, and the third angle BAC to the third angle EDF.

Next, let the fides which are oppofite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this cafe, the other fides fhall be equal, AC to DF, and BC to EF; and also the third angle BAC to the third EDF.

For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is

equal

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are oppofite; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and oppofite angle BCA; which is impoffible b; wherefore BC is b 16. 1. not unequal to EF, that is, it is equal to it; and AB is equal to_DE; therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D.

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PROP. XXVII, THEOR.

Fa ftraight line falling upon two other straight lines, makes the alternate angles equal to one another, these two straight lines shall be parallel.

Let the ftraight line EF, which falls upon the two flraight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

For, if it be not parallel, AB and CD being produced, fhall meet either towards B, D, or towards A, C; let them be produced and meet towards B, D in the point G; therefore GEF

is a triangle, and its exterior angle AEF is greater a than the a 16. 1, interior and oppofite angle

EFG; but it is alfo equal A I
to it, which is impoffible;
therefore AB and CD be-

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B

D

ftrated, that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel b to one another. AB therefore is parallel to CD. b 35. Def Wherefore, if a straight line, &c. Q. E. D.

PROP.

BOOK I:

IF

PROP. XXVIII. THEOR.

F a firaight line falling upon two other ftraight lines makes the exterior angle equal to the interior and oppofite upon the fame fide of the line; or makes the interior angles upon the fame fide together equal to two right angles; the two ftraight lines fhall be parallel to one another.

Let the ftraight line EF, which falls upon the two ftraight lines AB, CD, make the exterior angle EGB equal to the interior and oppofite angle GHD upon the fame fide; or make the interior angles on the fame fide BGH,

E

G

B

A

C

H

D

GHD together equal to two right angles; AB is parallel to CD.

Because the angle EGB is equal to the angle GHD, and the a 15. 1. angle EGB equal a to the angle AGH, the angle AGH is equal b. 1. to the angle GHD; and they are the alternate angles; therefore c By Hyp. AB is parallel b to CD. Again, because the angles BGH, GHD are equal e to two right angles, and that AGH, BGH are alfo d 13. 1. equal to two right angles; the angles AGH, BGH are equal to the angles BGH, GHD: Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a ftraight line, &c. Q. E. D.

See the

notes on

this propo

fition.

IF

PROP. XXIX. THEOR.

Fa ftraight line fall upon two parallel ftraight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and oppofite upon the fame fide; and likewife the two interior angles upon the fame fide together equal to two right angles.

Let the ftraight line EF fall upon the parallel ftraight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and oppofite, upon the fame fide, GHD; and the two interior angles BGH, GHD, upon the fame fide, are together equal to two right angles.

For,

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