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Then the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC: and the three interior angles ABC, BCA, ČAB shall be equal to two right angles.

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

Demonstration. Then, because CE is parallel to BA, and AC meets them, therefore

1. The angle ACE is equal to the alternate angle BAC (I. 29). Again, because CE is parallel to AB, and BD falls upon them, therefore

2. The exterior angle ECD is equal to the interior and opposite

angle ABC (I. 29) ; but the angle ACE was shown to be equal to the angle BAC, therefore

3. 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

4. The angles ACD and ACB are equal to the three angles

ABC, BAC, and ACB (Ax. 2); but the angles ACD, ACB are equal to two right angles (I. 13), therefore also

5. 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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Demonstration. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a poirt 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,

D

E

B

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 (I. 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.

D

B

Demonstration. 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 33.-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.

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Construction. Join BC.

Demonstration. Then, because AB is parallel to CD, and BC meets them, therefore

1. 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

2. The

base AC is equal to the base BD (I. 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

3. 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

4. AC is parallel to BD (I. 27), and AC was shown to be equal to BD.

Therefore, straight lines which, &c. Q.E.D.

PROPOSITION 34.-Theorem,

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them 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 BC shall bisect it.

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Demonstration. Because AB is parallel to CD, and BC meets them, therefore

1. The angle ABC is equal to the alternate angle BCD (I. 29). And because AC is parallel to BD, and BC meets them, therefore

2. 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 (I. 26), namely,

3. The side AB is equal to the side CD, and AC to BD, and

the angle BAC to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore

4. The whole angle ABD is equal to the whole angle ACD

(Ax. 2), and the angle BAC has been shown 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 BCT); therefore

5. The triangle ABC is equal to the triangle BCD (I. 4), and the diameter BC divides the parallelogram ACDB into two equal parts.

Q.E.D.

PROPOSITION 35.-Theorem.

Parallelograms upon the same base, and between the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC.

Then the parallelogram ABCD shall be equal to the parallelogram EBCF.

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Demonstration. If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base , be terminated in the same point D, then it is plain that

1. Each of the parallelograms ABCD, DBCF is double of the

triangle BDC (I. 34), and therefore

2. The parallelogram ABCD is equal to the parallelogram

DBCF (Ax. 6). But if the sides AD, EF, opposite to the base BC, be not terminated in the same point ; then, because ABCD is a parallelogram, therefore AD is equal to BC (I. 34); and for a similar reason, EF is equal to BC; wherefore

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