Book I.

~

I

F

PROP. XXVII. THEOR.

a straight 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 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 opposite angle

a. 16. 1.

b. 35. Def. ther way tho' produced ever so far are parallel to one another. AB therefore is parallel to CD. wherefore if a straight line, &c. Q. E. D.

IF

PROP, XXVIII. THEOR.

a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the fame fide of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.

E

G

B

Let the straight line EF which falls upon the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD upon the fame fide; or make the interior angles on the same side BGH, GHD together equal to two right angles. C AB is parallel to CD.

Because the angle EGB is equal to the angle GHD, and the

A

D

H

F

angle

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 m Parallel b 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 b. 27. 1. right angles; the angles AGH, BGH are equal to с. Ву Нур. the angles BGH, d. 1. 1. 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.

IF

PROP. XΧΙΧ. ΤΗHEOR.

a straight line falls upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same fide; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight 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 same side are together equal to two right angles.

E

See the notes on this Proposition.

AG

B

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

cause the angle AGH is greater than thể angle GHD, add to each of them the

F

angle BGH; therefore the angles AGH, PGH are greater than the angles BGH, GHD. but the angles AGH, BGH are equal to two righta. 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 same side less than two right angles, do meet * together if continually produced; therefore the . straight lines AB, CD if produced far enough shall meet. but they see the never meet, since 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 Proposition. 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 these the angle BGH,

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.

PROP. XXX. THEOR.

STRAIGHT lines which

are parallel to the same 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,

b. 27. 1. gles; therefore AB is parallel to CD. wherefore straight lines

&c. Q. E. D.

PROP. XXXI. PROB.

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

Let A be the given point, and BC the given straight line; it is

required to draw a straight line thro' E
the point A, parallel to the straight

AF

line BC.

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

D

C

2. 23. 1. line AD make 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 to one 6. 27. 1. another, EF is parallel to BC. therefore the straight line EAF is

drawn thro' the given point A parallel to the given straight line BC. Book I. Which was to be done.

PROP. XXXII. THEOR.

IF a fide of any triangle be produced, the exterior angle is equal to the two interior and opposite 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 opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB are together equal to two right angles.

Thro' the point C draw CE parallel to the straight line AB. and 2. 31. 1. because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal b. again because AB is parallel to CE, and BD falls upon them, the exterior angle

A

E b. 19. 1.

1

ECD is equal to the interior

B

and oppofite angle ABC. but

C

D

the angle ACE was shewn 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 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; C. 13. 1.
therefore also the angles CBA, BAC, ACB are equal to two right
angles. wherefore if a fide of a triangle &c. Q. E. D.

Book I. angles. And, by the preceeding Proposition, all the angles of there triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure. and the fame angles are equal to the angles of the figure, together with the angles at the 2. 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.

15. 1.

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 6.1.1. 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 D

is, by the foregoing Corollary,

A

C

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.

PROP. XXXIII. THEOR.

THE straight lines which join the extremities of tw equal and parallel straight lines, towards the same 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

B

rallel to CD, and BC meets them;
the alternate angles ABC, BCD

C

D

2.29. 1. 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 base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other anglest,