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PROPOSITION XXIX. THEOREM.

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 AB, CD. Then the alternate angles AGH, GHD shall be equal to one another ; the exterior angle EGB shall be equal to the interior and opposite

angle GHĎ upon the same side of the line EF; and the two interior angles BGH, GHD upon the same side of EF

shall be together equal to two right angles.

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First. For, if the angle AGH be not equal to the alternate angle

GHD, one of them must be greater than the other;

if possible, let AGH be greater than GHD, then because the angle AGH is greater than the angle GHD,

add to each of these unequals the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH,

GHD; (ax. 4.) but the angles AGH, BGH are equal to two right angles; (I. 13.) 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 two interior angles on the same side less than two right angles, will meet together if continually produced; (ax. 12.)

therefore the straight lines AB, CD, if produced far enough, will

meet towards B, D; but they never meet, since they are parallel by the hypothesis ;

therefore the angle AGH is not unequal to the angle GHD,

that is, the angle AGH is equal to the alternate angle GHD. Secondly. Because the angle AGH is equal to the angle EGB, (1. 15.)

and the angle AGH is equal to the angle GHD, therefore the exterior angle EGB is equal to the interior and opposite

angle GHD, on the same side of the line. Thirdly. Because the angle EGB is equal to the angle GHD,

add to each of them the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD;

(ax. 2.)

but EGB, BGH are equal to two right angles; (I. 13.) therefore also the two interior angles BGH, GHD on the same side of the line are equal to two right angles. (ax. 1.)

Wherefore, if a straight line, &c. Q.E.D.

PROPOSITION XXX. THEOREM. Straight lines which are parallel to the same straight line are parallel to each other.

Let the straight lines AB, CD, be each of them parallel to EF. Then shall AB be also parallel to CD.

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H

Let the straight line GHK cut AB, EF, CD. Then because GHK cuts the parallel straight lines AB, EF, in G, H: therefore the angle AGH is equal to the alternate angle GHF. (I. 29.)

Again, because GHK cuts the parallel straight lines EF, CD, in H, K; therefore the exterior angle GHF is equal to the interior angle HKD; and it was shewn that the angle AGH is equal to the angle GHF; therefore the angle AGH is equal to the angle GKD;

and these are alternate angles;

therefore AB is parallel to CD. (1. 27.). Wherefore, straight lines which are parallel, &c. Q. E. D,

PROPOSITION XXXI. PROBLEM. To draw a straight line through 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, through the point A, a straight line parallel to the straight line BC.

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A

A

F

Bóc
In the line BC take any point D, and join AD;

at the point A in the straight line AD, make the angle DAE equal to the angle ADC, (I. 23.) on the opposite side of AD;

and produce the straight line EA to F.

Then EF shall be parallel to BC. Because the straight line AD meets the two straight lines EF, BC, and makes the alternate angles EAD, ADC, equal to one another,

therefore EF is parallel to BC. (I. 27.) Wherefore, through the given point A, has been drawn a straight line EAF parallel to the given straight line BC. Q.E.F.

PROPOSITION XXXII. THEOREM. 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 together equal to two right angles.

Let ABC be a triangle, and let one of its sides B C be produced to D.

Then the exterior angle ACD shall be equal to the two interior and opposite angles CAŽ, ABC:

and the three interior angles ABC, BCA, CAB shall be equal to two right angles.

D

Through the point C draw CE parallel to the side BA. (1. 31.)

Then because CE is parallel to BA, and AC meets them, therefore the angle ACE is equal to the alternate angle BAC. (1. 29.)

Again, because CE is parallel to AB, and BD falls upon them, therefore the exterior angle ECD is equal to the interior and op

posite angle ABC; (1. 29.) but 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 opposite angles CAB, ABC. (ax. 2.) Again, because the angle ACD is equal to the two angles ABC, BAC,

to each of these equals add the angle ACB, therefore the angles ACD and ACB are equal to the three angles

ABC, BAČ, and ACB. (ax. 2.) but the angles ACD, ACB are equal to two right angles, (1. 13.) therefore also the angles ABC, BAC, ACB are equal to two right

angles. (ax. 1.)
Wherefore, if a side of any triangle be produced, &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 has sides.

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в For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.

Then, because the three interior angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides, therefore all the angles of these triangles are equal to twice as many

right angles as the figure has sides; but the same angles of these triangles are equal to the interior angles

of the figure together with the angles at the point F:

and the angles at the point F, which is the common vertex of all

the triangles, are equal to four right angles, (1. 15. Cor. 2.) therefore the same angles of these triangles are equal to the angles

of the figure together with four right angles ; but it has been proved that the angles of the triangles are equal to

twice as many right angles as the figure has sides; therefore all the angles of the figure together with four right angles,

are equal to twice as many right angles as the figure has sides. COR. 2. “All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles.

DB

Since every interior angle ABC with its adjacent exterior angle ABD, is equal to two right angles, (I. 13.)

therefore all the interior angles, together with all the exterior angles, are equal to twice as many right angles as the figure has sides ;

but it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides;

therefore all the interior angles together with all the exterior angles, are equal to all the interior angles and four right angles, (ax. 1.)

take from these equals all the interior angles, therefore all the exterior angles of the figure are equal to four right angles. (ax. 3.)

PROPOSITION XXXIII. THEOREM. The straight lines which join the extremities of two 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 same parts by the straight lines AC, BD.

Then AC, BD shall be equal and parallel.

Join BC. Then because AB is parallel to CD, and BC meets them, therefore the angle ABC is equal to the alternate angle BCD; (I. 29.)

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, (1. 4.)

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 because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another;

therefore AC is parallel to BD; (1. 27.)

and AC was shewn to be equal to BD. Therefore, straight lines which, &c. Q. E. D.

PROPOSITION XXXIV. THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts.

Let ACDB be a parallelogram, of which BC is a diameter.

Then the opposite sides and angles of the figure shall be equal to one another; and the diameter B Č shall bisect it.

B

Because AB is parallel to CD, and BC meets them, therefore the angle ABC is equal to the alternate angle BCD. (1. 29.)

And because AC is parallel to BD, and BC meets them, therefore the angle ACB is equal to the alternate angle CBD. (I. 29.)

Hence in the two triangles ABC, CBD, because the two angles ABC, BCA in the one, are equal to the two

angles BCD, CBD in the other, each to each; and one side BC, which is adjacent to their equal angles, common to

the two triangles ; therefore their other sides are equal, each to each, and the third angle

of the one to the third angle of the other, (1. 26.) namely, the side AB to the side CD, and AC to BD, and the angle

BAC to the angle B.DC.
And because the angle ABC is equal to the angle BCD,

and the angle CBD to the angle ACB,
therefore the whole angle ABD is equal to the whole angle ACD;

(ax. 2.)

and the angle BAC has been shewn to be equal to BDC'; therefore the opposite sides and angles of a parallelogram are equal to

one another. Also the diameter BC bisects it. For since AB is equal to CD, and BC common, the two sides AB,

BC, are equal to the two DC, CB, each to each, and the angle ABC has been proved to be equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD; (1. 4.) and the diameter B C divides the parallelogram ACDB into two equal parts.

Q.E.D.

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