Book the other fides fhall be equal, each to each, viz. AB to DE, and AC to DF, and the third angle BAC to the third angle EDF.

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 equal a4. 1. to the angle DEF; therefore the bafe GC is equal a to the bafe 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 lefs to the greater, which is impaffible; 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.

#

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 to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore the bafe AH is equal to the bafe DF, and the triangle ABH to the triangle DEF, and the other angles fhall be equal, each to each, to which the equal fides are oppofite; therefore the angle BHA is equal to the angle EFD;

EFD; but EFD is equal to the angle BCA; therefore alfo Book I. the angle BHA is equal to the angle BCA, that is, the exte-~~ rior angle BHA of the triangle AHC is equal to its interior and oppofite angle BCA; which is impoffible b; wherefore b 16. 1. BC is 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 bafe AC is equal to the bafe DF, and the third angle BAC to the third angle EDF. Therefore, if two thi angles, &c. Q. E. D.

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 straight line EF, which falls upon the two straight 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 shall 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. I. interior and oppofite

angle EFG; but it is alfo equal to it, which A is impoffible; therefore,

AB and CD being pro

manner it may be de

G

D

monftrated that they do not meet towards A, C ; but those ftraight lines which meet neither way, though produced ever

fo far, are parallel b to one another. AB therefore is parallel b 30. Def. to CD. Wherefore, if a straight line, &c. Q. E. D.

PROB

Book I.

b 27. I.

PROP. XXVIII. THEOR.

Fa ftraight 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 straight 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, A
GHD together equal to two
right angles; AB is parallel
to CD.

E

C

H

D

F

Because the angle EGB is equal to the angle GHD, a 15. 1. and the angle EGB equal a to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel b to c By Hyp. CD. Again, because the angles BGH, GHD are equal to d 13. 1. two right angles; and that AGH, BGH, are also equald 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 N.

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 likewise the two interior angles upon the fame fide together equal to two right angles.

Let

Let the ftraight line EF fall upon the parallel straight lines Book I. 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.

A.

E

G

a 27. I.

B

D

H

F

For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be parallel to CD a; but AB is alfo parallel to CD;, therefore-two-ftraight, lines) are drawn through the K fame point G, parallel to CD, and yet not coinciding with one another, which is im poffible b. The angles AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGHc; and AGH is proved to be equal to GHD; therefore EGB is likewife equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equald to d 13.1. two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D.

COR. If two lines KL and CD make, with EF, the two angles KGH, GHC together lefs than two right angles, KG and GH will meet on the fide of EF on which the two angles are that are less than two right angles.

For, if not, KL and CD are either parallel or they meet on the other fide of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet toward the points L and D; for the angles LGH, GHD would then be two angles of a triangle, and lefs than two right angles; but this is impoffible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles d, of which the two KGH, CHG are by fuppofition less than two right angles; therefore the other two, HGL, GHD are greater than two right angles. Therefore, fince KL and CD are not parallel, and do not meet towards L and D, they will meet if produced towards K and C.

PROP.

b 11. Ax.

C 15. I.

Book I.

PROP. XXX.

THEOR.

TRAIGHT lines which are parallel to the fame

STRAIGHT ile er "parcalled 'te another.

Let AB, CD be each of them parallel to EF; AB is alfo parallel to CD.

Let the ftraight line GHK cut AB, EF, CD; and because GHK cuts the parallel ftraight lines AB, EF, the angle 19. 1. AGH is equal a to the angle

GHF. Again, because the
ftraight line GK cuts the
parallel straight lines EF,
CD, the angle GHF is equai
a to the angle GKD; and it
was fhewn that the angle
AGK is equal to the angle
GHF; therefore also AGK

G

B

is equal to GKD; and they are alternate angles; therefore. 27.1. AB is parallel b to CD. Wherefore ftraight lines, &c. Q E. D.

T

O draw a ftraight line through a given point parallel to a given ftraight line.

Let A be the given point, and BC the given straight line; it is required to draw a

ftraight line through the
point A, parallel to the
ftraight line BC.

In BC take any point

D, and join AD; and at

A

F

C

a 23. 1. the point A, in the straight line AD make a the angle DAE equal to the angle ADC; and produce the straight line EA to F.

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal b 27. 1. to one another, EF is parallelb to BC. Therefore the straight