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

35 Def.

PROPOSITION XXVIII.

Theor.-If a straight line, falling upon two other straight

lines, make the exterior angle equal to the interior and opposite upon the same side of the line, or make the interior angles upon the same side together equal to two right angles, the two straight lines shall be parallel to one another.

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+ Hyp.

15. 1.

tl Ax.

Let the straight line EF, which falls upon the two straight
lines AB, CD, make the exterior angle
EGB equal so the interior and opposite

E
angle GHD upon the same side ; or make
the interior angles on the same side, BGH,

G
A-


GID together equal to two right angles :

D AB shall be parallel to CD.

Because the angle EGB is equal to the angle GHD, and the angle EGB is equal * to the angle AGH, therefore 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 are equal * to two right angles, and that AGH, BGH are also equal * to two right angles, therefore the angles AGII, 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 straight line, &c. Q. E. D.

* 27.1. * Hyp:

13. 1. + 1 Ax. + 3 Ax.

† 27. 1.

PROPOSITION XXIX.

See the notes on this

proposition.

Thron.-If a straight line fall 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 side ; 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

F

AB, CD : the alternate angles AGH, GAD shall be equal to one
another; and the exterior angle EGB shall be equal to the in-
terior and opposite, upon the same side,
GHD; and the two interior angles BGH,

E GHD upon the same side, shall be together equal to two right angles.

B For if AGH be not equal to GHD, one

D of them must be greater than the other ; let AGH be the greater; and because the angle Agh is greater than the angle GHD, add to each of them the angle BGH ; therefore, the angles AGH, BGH are greater † than the angles BGH, GHD: but the angles † 4 Ax. AGH, BGH are equal to two right angles ; therefore the • 13. 1. 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, will meet * together if continually produced; • 12 A therefore the straight lines AB, CD, if produced far enough, See the will meet : but they never meet, since they are parallel by the this propohypothesis ; therefore the angle Agh is not unequal to the sition. angle GHD, that is, it is equal to it: but the angle AGH is equal * to the angle EGB ; therefore likewise EGB is equal + • 15. 1. to GHD: add to each of these the angle BGH; therefore the 1 Ax. angles EGB, BGH are equal † to the angles BGH, GHD: but + 2 Ax. EGB, BGH are equal * to two right angles; therefore also . 13. 1. BGH, GHD are equal + to two right angles. Wherefore, if a t 1 Ax. straight line, &c. Q. E. D.

PROPOSITION XXX.

TAEOR.--Straight lines which are parallel to the same straight

line, are parallel to each other. Let AB, CD be each of them parallel to EF: AB shall be parallel to CD.

Let the straight line GHK cut AB, EF, CD: and because Ghi cuts the parallel

-B straight lines AB, EF, the angle AGH is Н

E

F equal* to the angle GHF. Again, because C

K.

D • 29. I. the straight line GK cuts the parallel straight lines EF, CD, the angle GAF is equal to the angle . 29. 1. GKD: and it was shewn, that the angle AGK is equal to the angle

A

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GHF; therefore also AGK is equal + to GKD: and they are alternate angles ; therefore AB is parallel * to CD. Wherefore, straight lines, &c.

Q. E. D.

PROPOSITION XXXI.

PROB.--To draw a straight line, through a given point, pa

rallel to a given straight line.

# 23. 1.

Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight line BC. In BC take any point D, and join AD;

E A F and at the point A, in the straight line AD, make the angle DAE equal to the angle ADC; and produce the straight line EA to

B D) С F: EF shall be parallel to BC.

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel * to BC. Therefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Which was to be done.

* 27. 1.

PROPOSITION XXXII.

THEOR.If a side 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 sides BC be produced to D; the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB shall together be equal to two right angles.

Through the point c, draw CE parallel * to the straight line AB: and

E because AB is parallel to CE, and AC meets them, the alternate angles BAC,

с ACE are equal. Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal + to the interior and opposite angle

* 31. 1.

B

* 29, 1.

† 29. 1.

ABC: but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACR is equal + + 2 Ax. to the two interior and opposite angles CAB, ABC: to each of these equals add the angle ACB ; and the angles ACD, ACB are equal + to the three angles CBA, BAC, ACB ; but the + 2 Ax. angles ACD, ACB are equal * to two right angles ; therefore • 13. 1. also the angles CBA, BAC, ACB are equal t to two right + 1 Ax. angles. Wherefore, if a side of any triangle, &c. Q. E. D.

Cor. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure E has sides. For

any rectilineal figure ABCDE, can be divided into as many triangles as the figure

A has sides, by drawing straight lines from a point F within the figure, to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure : and the same 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 *, together with four right angles. Therefore all the * 2 Cor, angles of the figure, together with four right angles, are equal 15. I. to twice as many right angles as the figure has sides.

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 ABD, is equal* to two right angles, there

# 13. I. fore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure ; that is, by the fore- D going corollary, 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.

B

PROPOSITION XXXIII.

THEOR.The straight lines which join the extremities of two

equal and parallel straight lines towards the same parts, are also themselves equal and parallel.

. 29. 1.

Let AB, CD be equal and parallel straight lines, and joined
towards the same parts by the straight lines AC, BD: AC,
BD shall be equal and parallel.
Join BC: and because AB is parallel to

B
CD, and BC meets them, the alternate an-
gles * ABC, BCD are equal: and because AB
is equal to CD, and BC common to the two
triangles ABC, DCB, the two sides AB, BC
are equal to the two DC, CB, each to each : and the angle
ABC was proved to be 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 to each, to which the equal sides are opposite :
therefore the angle ACB is equal to the angle CBD: and be-
cause the straight line BC meets the two straight lines AC,
BD, and makes the alternate angles ACB, CBD equal to one
another, AC is parallel * to BD; and it was shewn to be equal
to it. Therefore, straight lines, &c. Q. E. D.

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PROPOSITION XXXIV.

i 29.1.

THEOR.---The opposite sides and angles of paralellograms are

equal to one another, and the diameter bisects them, that is, divides them into two equal parts.

N.B. A parallelogram is a four-sided figure, of which the opposite sides are parallel : and the diameter is the straight line joining two of its opposite angles.

Let ACDB be a parallelogram, of which BC is a diameter: the opposite sides and angles of the figure shall be equal to one another, and the diameter BC shall bisect it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD A

B are equal* to one another : and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal * to one another : wherefore the two triangles ABC, CBD have two angles ABC, BCA in the one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides are equal, each to each, and the third angle of the one, to the third angle of the other *, viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC: and because the angle ABC is equal

29. ).

. 29. 1.

26.1.

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