(Edited by Dionysius Lardner, 1855)

Proposition XV. Theorem.

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

(87) | If two right lines (A B and C D) intersect one another, the vertical angles are equal (C E A to B E D, and C E B to A E D). |

Because the right line C E stands upon the right line A B, the angle A E C together with the angle C E B is equal to two right angles (XIII); and because the right line B E stands on the right line C D, the angle C E B together with the angle B E D is equal to two right angles (XIII); therefore A E C and C E B together are equal to C E B and B E D; take away the common angle C E B, and the remaining angle A E C is equal to B E D.

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

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Previous: Proposition 14

This proposition in other editions:

This proof may shortly be expressed by saying, that opposite angles are equal, because they have a common

supplement(84).It is evident that angles which have a common supplement or complement (85) are equal, and that if they be equal, their supplements and complements must also be equal.

(88) The

converseof this proposition may easily be proved, scil. If four lines meet at a point, and the angles vertically opposite be equal, each alternate pair of lines will be in the same right line. For if C E A be equal to B E D, and also C E B to A E D, it follows that C E A and C E B together are equal to B E D and A E D together. But all the four are together equal to four right angles (83), and therefore C E A and C E B are together equal to two right angles, therefore (XIV) A E and A B are in one continued line. In like manner it may be proved, that C E and D E are in one line.