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which is impossible. Consequently the above supposition is false; MN is not parallel to Ef. In like manner it can be proved that no other right line through the point B is parallel to EF, except CD. Hence CD is parallel to EF. This, &c. [See NOTE K.]

ART. 16. If a right line intersect two right lines, and make the external angle equal to the farther internal angle at the same side of the intersecting line, these two latter right lines are parallel.

Let AB be the right line intersecting two other lines EF, E CD, and making the external angle KAF equal to the farther internal one ABD. Then EF is parallel to CD.

C

B

K

A

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DEM. BY ART. 11, the angles KAF and EAB are equal; therefore, since KAF is equal to ABD, also EAB is equal to ABD Hence, by ART. preceding, EF is parallel to CD. This, &c.

ART. 17. If a right line intersect two right lines, and make the two internal angles at the same side of the intersecting line together equal to two right angles, these two latter right lines are parallel.

In a figure similar to the preceding, let the two internal angles FAB and ABD be together equal to two right angles. Then EF is parallel to CD.

DEM. BY ART. 9, the angles FAB, EAB, together, are equal to two right angles; therefore they are equal to the angles FAB, ABD, together *. Take away from these equals the common angle FAB, and EAB remains equal to ABD †. Hence, by ART. 15, EF is parallel to CD. This, &c.

ART. 18. If a right line intersect two parallel right lines, and another right line be drawn parallel to the inter

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sector from any point in either of the parallels, it will meet the other, if produced sufficiently; and its length between the parallels will be equal to the length of the intersector between the parallels.

Let AB be the right line intersecting the two parallel right lines EF, CD. Then if from any point, G, in either of these parallels, EF, a right line be drawn parallel to AB, this right line will meet the other parallel, CD; and its

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intercepted part, between the parallels EF and CD, will be equal to the intercepted part AB.

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Since EF is

DEM. Take BI equal to AG, and join IG. parallel to CD, and the points A and G on the one are equally remote as the points в and 1 on the other, therefore, by DEF. IX. the lines AB and GI are equal. Join AI. In the triangles IBA, AGI, the side AI is common, the side BI has been taken equal to the side AG, and the side AB has been proved equal to the side GI; consequently, by ART. 6, the angle IAB, opposite to the side BI, is equal to the angle AIG, opposite to the corresponding side AG. But these are alternate angles; therefore, by ART. 15, the lines AB and GI are parallel; and therefore, by ART. 13, the two internal angles BAG, AGI are together equal to two right angles. Now the right line drawn from the point G parallel to AB must fall on the line GI. For suppose it to fall in any other direction, as GH; the two internal angles BAG, AGH, would, by ART. 13, be together equal to two right angles, and, therefore, equal to the angles BAG, AGI, together, which have been proved also equal to two right angles. Consequently, taking away from both the common angle BAG, there would remain the angle AGI equal to the angle AGH, the less to the greater, which is impossible. Hence, the above supposition was false; the right line drawn from the point G, parallel to AB, does not fall in any other direction than GI, and its intercepted part, G1, has been proved equal to AB. This, &c. [See NOTE L.]

LESSON IV.

ART. 19. Right lines which join the adjacent extremities of two equal and parallel right lines are themselves equal and parallel.

Let AB, CD be two equal and parallel right lines, whose adjacent extremities are joined by the right lines AC, BD. Then AC, BD are also equal and parallel.

A

D

B

DEM. Join either pair of opposite extremities c and B. Now in the triangles ABC, BCD, the sides AB and CD are equal, the side BC is common, and the contained angles ABC, BCD, are equal by ART. 12. Hence, by ART. 1., AC is equal to BD; also the angles ACB, DBC, being equal by ART. 2, the lines AC, BD are paralleled by Art. 15. This, &c.

Observation. The equality of AC to BD follows immediately from DEF. IX.

DEF. XI. A parallelogram is a four-sided rectilineal figure, each pair of whose opposite sides are parallel.

Thus, if AC and BD are parallel, and

also AB and CD, the figure ABDC is called a parallelogram. AD and BC are called its diagonals.

B

ART. 20. The opposite sides of a parallelogram are equal.

Let ABDC be a parallelogram. Then

AB and CD (or AC and BD) are equal.

A

DEM. Inasmuch as the right line AC meets the parallel right lines AB and CD, and from the point в in one of these parallels a right line BD is drawn meeting the other CD, and parallel to AC, (by DEF. XI.,) therefore BD is equal to AC, by ART. 18. In the same way it is proved that AB is equal

to CD; the line AB being drawn parallel to CD from a point in AC which is parallel to BD. This, &c.

ART. 21. The opposite angles of a parallelogram are equal.

In the preceding figure the angles ABD and ACD (or CDB and BAC) are equal.

DEM. BY ART. 13, ABD and BDC together are equal to two right angles; and therefore to BDC and DCA together, which are by the same ART. also equal to two right angles. Hence, taking away BDC from both, ABD remains equal In the same manner BAC is proved equal to CDB.

to DCA. This, &c.

ART. 22. A parallelogram is divided into two equal parts by its diagonal.

In the parallelogram ABDC, let the diagonal BC be drawn. Then the triangle ABC is equal to the triangle BDC.

C

A

D

B

DEM. BY ART. 20, AB equals CD, and AC equals BD; also the side BC is common to the two triangles ABC, BDC. Hence, by ART. 7, these triangles are equal. This, &c.

ART. 23. Parallelograms on the same base, and between the same parallels, are equal.

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Then these pa

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Let ABCD, EBCF, be two parallelograms on the same base BC, and between the same parallels BG, af. rallelograms are equal.. DEM. BY ART. 14, the angles ABC, DCG, are equal; and also the angles EBC, FCG. Therefore, taking EBC from ABC, and FCG from DCG, the angle ABE remains equal to the angle DCF. But, by ART. 20, AB is equal to DC, and EB to FC; consequently the triangles ABE, DCF, are equal, by ART. A 3. Now, if from the four-sided figure ABCF we take the triangle ABE, the parallelogram EBCF remains; and if from the same figure we take the equal triangle DCF, the parallelogram ABCD remains. Hence, as from the same quantity we successively take the equal triangles ABE, DCF,

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F

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the remainder in one case will be equal to the remainder in the other; that is, the parallelogram EBCF to the parallelogram ABCD. This, &c.

ART. 24. Parallelograms on equal bases and between the same parallels are equal.

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By ART. 20, CD is

DEM. Draw the right lines AC, BD. equal to GH, and therefore to AB, by the terms of the present ART. Consequently, by ART. 19, AC is equal and parallel to BD; and the figure ABDC is a parallelogram by DEF. XI. Now, by preceding ART. EABF is equal CABD; and by the same ART. CGHD is also equal to CABD. Hence, EABF is equal to CGHD. This, &c.

ART. 25. Triangles upon the same base and between the same parallels are equal.

Let ABC, AEC, be two triangles

B E

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DEM. Draw AD parallel to CB, and CF parallel to AE. By ART. 23, the parallelograms ADBC, AEFC, are equal. Hence the triangles ABC, AEC, which by ART. 22 are their halves, are also equal *. This, &c.

ART. 26. Triangles upon same parallels are equal.

Let ABC, DEF, be two triangles upon equal bases AC, DF, and between the same parallels AF, BE.

triangles are equal.

equal bases and between the

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Then these

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* Ax. 7. The halves of equal things are equal.

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