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Book I.

a. 16. I.

PROP. XXVII. THEOR.

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a ftraight line falling upon two other straight lines makes the alternate angles equal to one another, these two ftraight lines fhall 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 BD or towards AC. let them be produced and meet towards BD in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater than the interior and oppofite angle EFG; but it is alfo equal to

it, which is impoffible. there

a

fore AB and CD being pro- A

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b. 35. Def. ther way tho' 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.

Ima free, or angle equal to the interior and op

F a straight line falling upon two other straight lines

op

pofite 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 straight line EF which falls upon the two straight lines
AB, CD make the exterior angle

EGB equal to the interior and op-
pofite angle GHD upon the fame
fide; or make the interior angles
on the fame fide BGH, GHD to-

gether equal to two right angles. C
AB is parallel to CD.

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

E

G

-B

D

H

F

angle

b. 27. I.

c. By Hyp.

angle EGB equal to the angle AGH, the angle AGH is equal to Book I. the angle GHD; and they are the alternate angles; therefore AB is ne pwallel to CD. again, because the angles BGH, GHD are equal ca. 15, 1. t› two right angles, and that AGH, BGH are also 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.

IF

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d. 13. 1.

notes on this Propofition.

Fa ftraight line falls upon two parallel ftraight lines, See the 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 straight 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 opposite upon the fame fide, GHD; and the two interior angles BGH, GHD upon the fame fide are together equal to two right angles.

For if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater. and be

cause the angle AGH is greater than the

angle GHD, add to each of them the

E

A G

B

H

D

F

angle BGH; therefore the angles AGH, EGH are greater than the an

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gles BGH, GHD. but the angles AGH, BGH are equal to two right a. 13. 1. angles; therefore the angles BGH, GHD are less than two right angles. but those straight lines which with another straight 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 12. Ax. straight lines AB, CD if produced far enough shall meet. but they See the never meet, fince they are parallel by the Hypothesis. therefore the notes on this angle AGH is not unequal to the angle GHD, that is, it is equal to Propofition. it. but the angle AGH is equal to the angle EGB; therefore like-b. 15. 1. wife EGB is equal to GHD. add to each of thefe the angle BGH,

there

Book I therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to two right angles; therefore also c. 13. 1. BGH, GHD are equal to two right angles. wherefore if a straight line &c. Q. E. D.

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STRAIGHT 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 alfo parallel to CD.

Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, a. 29. 1. EF, the angle AGH is equal to

a

a

the angle GHF. again, because the
straight line GK cuts the parallel A-
ftraight lines EF, CD, the angle
GHF is equal to the angle GKD. E
and it was fhewn that the angle
AGK is equal to the angle GHF;
therefore alfo AGK is equal to

GKD. and they are alternate an

C

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b. 27. 1. gles; therefore AB is parallel to CD. wherefore ftraight lines &c. Q. E. D.

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O draw a ftraight line thro' a given point parallel to a given straight line.

E

A F

Let A be the given point, and BC the given ftraight line; it is
required to draw a straight line thro'
the point A, parallel to the straight
line BC.

In BC take any point D, and join
AD; and at the point A in the ftraight B

a. 23. 1. line AD make the angle DAE equal

D

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 to one

b. 27.1. another, EF is parallel to BC. therefore the straight line EAF is

drawn

drawn thro' the given point A parallel to the given straight line BC. Book I. 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 produced to D. the exterior angle ACD is equal to the two interior and oppofite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, ÇAB are together equal to two right angles.

Thro' the point C draw CE parallel to the straight line AB. and a. 31. I. because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal. again because AB is parallel to CE, and BD falls

A

E b. 19. 1.

C

D

upon them, the exterior angle
ECD is equal to the interior
B
and oppofite angle ABC. but
the angle ACE was fhewn to be equal to the angle BAC; therefore
the whole exterior angle ACD is equal to the two interior and op-
pofite angles CAB, ABC. to thefe equals add the angle ACB, and
the angles ACD, ACB are equal to the three angles CBA, BAC,

c

ACB. but the angles ACD, ACB are equal to two right angles; c. 13. 1.

therefore alfo the angles CBA, BAC, ACB are equal to two right angles. wherefore if a fide of a triangle &c. Q. E. D.

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Book I. angles. And, by the preceeding Propofition, all the angles of the ~ 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 a. 2. Cor. point F which is the common Vertex of the triangles; that is, 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.

35. 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 6.13.1. 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 is, by the foregoing Corollary,

D B

they are equal to all the interior angles of the figure, together with four right angles. therefore all the exterior angles are equal to four right angles.

PRO P. XXXIII. THEOR.

THE 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 parallel straight lines, and joined towards the fame parts by the straight A

lines AC, BD; AC, BD are alfo

equal and parallel.

Join BC, and because AB is pa

rallel to CD, and BC meets them;

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B

D

a. 29. 1. are equala; and because AB is equal to CD, and BC common tơ

4.4. I.

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 base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, each

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