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Book I.

a 16. I.

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IF a ftraight line falling upon two other ftraight line: makes the alternate angles equal to one another, these two straight lines fhall be parallel.

Let the ftraight line EF, which falls upon the two ftraight 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 interior and oppofite angle EFG; but it is alfo equal to it, which is impoffible; therefore AB and CD being produced do not meet towards B, D. In like manner it may be demonftrated that they do C

not meet towards A, C ; but

thofe ftraight lines which

meet neither way, though

A

E

B

G

F

D

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

PROP. XXVIII. THEOR.

IF a 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 straight 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 LGB equal to the interior and oppofite angle GHD upon the Afame fide; or make the interior angles on the fame fide BGH, GHD together equal to two right angles; AB is parallel to CD.

Because the angle EGB is e qual to the angle GHD, and the

E

G

B

-D

H

angle

a

Book I.

a 15. I.

13. 1.

angle EGB equal to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel to CD. Again, because the angles BGH, GHD b 27. 1. are equal to two right angles, and that AGH, BGH are alfo By Hyp. equal to two right angles; the angles AGH, BGH are equald 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 straight line, &c. Q. E. D.

I

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this propo

a ftraight line falls upon two parallel ftraight lines, it See the makes the alternate angles equal to one another; and notes on the exterior angle equal to the interior and oppofite upon fition, 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 E angles BGH, GHD upon the fame fide are together equal to two right angles.

A G

For, if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater; and C because the angle AGH is greater than the angle GHD, add to each of

B

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them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles AGH, BGH are equal to two right angles; therefore the angles a 13. 1. BGH, GHD are lefs than two right angles; but thofe ftraight lines which, with another ftraight line falling upon them, make the interior angles on the fame fide lefs than two right angles, do meet together if continually produced; therefore the r2. án. ftraight lines AB, CD, if produced far enough, fhall meet; but See the they never meet, fince they are parallel by the hypothefis; this propotherefore the angle AGH is not unequal to the angle GHD, fition. that is, it is equal to it; but the angle AGH is equal to the b 15. 1. angle EGB; therefore likewife EGB is equal to GHD; add to

C 2

each

notes on

€ 13. I.

Book I. each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D.

a 29. 1.

b 27. 1.

# 23. I.

b 27. 1.

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TRAIGHT lines which are parallel to the fame straight line are parallel to one another.

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

Let the ftraight line GHK cut AB, EF, CD; and because GHK cuts the parallel ftraight

a

E

lines AB, EF, the angle AGH
is equal to the angle GH".
Again, because the ftraight line A-
GK cuts the parallel straight lines
EF. CD, the angle GHF is equal
to the angle GKD; and it was
fhewn that the angle AGK is e-
qual to the angle GHF; there
fore alfo AGK is qual to GKD;
and they are alternate angles;

C

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therefore AB is parallel to CD. Wherefore ftraight lines, &c. Q. E. D.

T

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O draw a ftraight line through a given point paral. lel to a given straight 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 E

ftraight line BC.

a

In BC take any point D, and jo'n AD; and at the point A in the ftraight line AD make the ngle B DAE equal to the angle ADC; and produce the ftraight line EA to F

D

A F

C

Because the straight line AD, which meets the two straight lines BC, LF, makes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. Therefore the ftraight line

EAF

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EAF is drawn through the given point A parallel to the given Book I. ftraight line BC. Which was to be done.

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IF a fide of any triangle be produced, the exterior angle
is equal to the two interior and oppofite angles; and
the three interior angles of every triangle are equal to
two right angles.

Let ABC be a triangle, and let one of its fides BC be pro-
duced to D; the exterior angle ACD is equal to the two inte
rior and oppofite angles CAB, ABC; and the three interior
angles of the triangle, viz. ABC, BCA, CAB are together e-
qual to two right angles.

Through the point C draw
CE parallel to the ftraight
line AB; and because AB is
parallel to CE and AC meets
them, the alternate angles
BAC, ACE are equal b. A-
gain, because AB is parallel
to CE, and BD falls upon
them, the exterior angle ECD B
is equal to the interior and

a 31. X.

A

E

b 29. I.

D

C

oppofite angle ABC; but the angle ACE was fhown to be equal
to the angle BAC; therefore the whole exterior angle ACD
is equal to the two interior and oppofite angles CAB, ABC;
to these equals add the angle ACB, and the angles ACD, ACB
are equal to the three angles CBA, BAC, ACB; but the angles
ACD, ACB are equal to two right angles; therefore alto the c 13. I.
angles CBA, BAC, ACB are equal to two right angles.
Wherefore, if a fide of a triangle, &c. Q. E. D.

с

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Book I. to each of its angles. And, by the preceding propofition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are fides of the figure; and the fame angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is a, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides.

a 2. Cor. 15. I.

COR. 2. All the exterior angles of any rectilineal figure, are together equal to four right angles.

Because every interior angle

ABC, with its adjacent exterior

b 13. I. ABD, is equal to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are fides of the figure; that

a 29. I.

is, by the foregoing corollary, D B
they are equal to all the inte-

rior angles of the figure, toge

ther with four right angles; therefore all the exterior angles are equal to four right angles.

THE

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HE ftraight lines which join the extremities of two equal and parallel ftraight lines, towards the fame

parts, are also themselves equal and parallel.

Let AB, CD be equal and pa-
A
rallel ftraight lines, and joined
towards the fame parts by the
ftraight lines AC, BD; AC, BD
are alfo equal and parallel.

Join BC; and because AB is pa-
rallel to CD, and BC meets them,
the alternate angles ABC, BCD

C

B

D

are equal; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two fides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the bafe AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles

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