(Edited by Dionysius Lardner, 1855)

Proposition XXIX. Theorem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(116) | If a right line (E F) intersect two parallel right lines (A B and C D), it makes the alternate angles equal (A G H to G H D, and C H G to H G B); and the external angle equal to the internal and opposite upon the same side (E G A to G H C, and E G B to G H D); and also the two internal angles at the same side (A G H and C H G, B G H and D H G) together equal to two right angles. |

1° The alternate angles A G H and G H D are equal; for if it be possible, let one of them A G H be greater than the other, and adding the angle B G H to both, A G H and B G H together are greater than B G H and G H D; but A G H and B G H together are equal to two right angles (XIII), therefore B G H and G H D are less than two right angles, and therefore the lines A B and C D, if produced, would meet at the side B D (Axiom 12); but they are parallel (hyp.), and therefore cannot meet, which is absurd. Therefore neither of the angles A G H and G H D is greater than the other; they are therefore equal.

In the same manner it can be demonstrated, that the angles B G H and G H C are equal.

2° The external angle E G B is equal to the internal G H D; for the angle E G B is equal to the angle A G H (XV); and A G H is equal to the alternate angle G H D (first part); therefore E G B is equal to G H D. In the same manner it can be demonstrated, that E G A and G H C are equal.

3° The internal angles at the same side B G H and G H D together are equal to two right angles; for since the alternate angles G H D and A G H are equal (first part), if the angle B G H be added to both, B G H and G H D together are equal to B G H and A G H and therefore are equal to two right angles (XIII). In the same manner it can be demonstrated, that the angles A G H and G H C together are equal to two right angles.

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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This proposition in other editions:

(117) Cor. 1.—If two right lines which intersect each other (A B, C D) be parallel respectively to two others (E F, G H), the angles included by those lines will be equal.

Let the line I K be drawn joining the points of intersection. The angles C I K and I K H are equal, being alternate; and the angles A I K and I K F are equal, for the same reason. Taking the former from the latter, the angles A I C and H K F remain equal. It is evident that their supplements C I B and G K F are also equal.

(118) Cor. 2.—If a line be perpendicular to one of two parallel lines, it will be also perpendicular to the other; for the alternate angles must be equal.

(119) Cor. 3.—The parts of all perpendiculars to two parallel lines intercepted between them are equal.

For let A B be drawn. The angles B A C and A B D are equal, being alternate; and the angles B A D and A B C are equal, for the same reason; the side A B being common to the two triangles, the sides A C and B D must be equal (XXVI).

(120) Cor. 4.—If two angles be equal (A B C and D E F), and the sides A B and D E be parallel, and the other sides B C and E F lie at the same side of them, they will also be parallel; for draw B E. Since A B and D E are parallel, the angles G B A and G E D are equal. But, by hypothesis, the angles A B C and D E F are equal; adding these to the former, the angles G B C and G E F are equal. Hence the lines B C and E F are parallel.