PROPOSITION XIII, THEOREM.-If a straight line (AB) standing upon another (CD) forms angles with it, they are either two right angles, or are together equal to two right angles. DEMONSTRATION. For if the line AB is perpendicular to CD, the angles ABC and ABD are two right angles (a). But if not, draw BE perpendicular to CD (b), and it is evident that the angles ABD and ABC are together equal to the angles EBD and EBC, and therefore to two right angles. B (a) Def. 9. (b) I. 11. D COROLLARY 1. From this proposition it is evident, that if several straight lines stand on the same side of another straight line at the same point, and make angles with it, all those angles are together equal to two right angles. COROLLARY 2. Also, if two straight lines intersect, the four angles which they form at the point of intersection are together equal to four right angles. COROLLARY 3. And consequently, if any number of straight lines diverge from the same point, all the angles which they make taken together are equal to four right angles. SCHOLIUM. It is necessary, in the enunciation of this proposition, to insert the words "forms angles with it," to exclude the case in which the line AB stands at either extremity of CD. PROPOSITION XIV. THEOREM. -If two straight lines (CB and BD) meet another straight bine (AB) at the same point and on opposite sides, and make the adjacent angles with it (ABC and ABD) together cqual to two right angles, those two straight lines (CB and BD) will form one continued straight line. B DEMONSTRATION. For, if possible, let BE, and not BD, be the continuation of the straight line CB. Then ABC and ABE are together equal to two right angles (a), and because ABC and ABD are together equal to two right angles (b), therefore ABC and ABE taken together are equal to ABC and ABD taken together (c). If now from these equals we take away the angle ABC, which is common to both, the remaining angles ABE and ABD shall be equal (d) a part to the whole, which is absurd; therefore (a) I. 13. BE is not the continuation of CB. And in like manner it may be proved that no other straight line except BD can be the continuation of CB; therefore BD and CB form one continued straight line. SCHOLIUM. The above proposition is proved by the "reductio ad absurdum." It is necessary that the two straight lines CB and BD should be on opposite sides of AB; for otherwise they might form angles with it together equal to two right angles, without being in the same continued straight line, as in the annexed figure. PROPOSITION XV. THEOREM.-If two straight lines (AB and CD) intersect one another, the vertical angles are equal (CEA to DEB, and CEB to AED). E DEMONSTRATION. Because the straight line AE forms with CD the angles CEA and AED, they are together equal to two right angles (a); also because the straight line DE forms with AB the angles AED and DEB, they are together equal to two right angles (a); therefore CEA and AED taken together are equal to AED and DEB taken together (b). Îf now from these equals we take away the angle AED, which is common to both, the remaining angles CEA and DEB shall be equal (c). And in like manner it can be proved that the angles CEB and AED are equal. (a) I. 13. (b) Ax. 1. (c) Ax. 3. SCHOLIA. 1. The second and third corollaries to proposition xiii. may also be drawn as corollaries from the present one. (a) Hypoth. (b) Ax. 2. 2. The converse of this proposition may be demonstrated as follows. If four straight lines meet in the same point, and make the vertical angles equal, each alternate pair of lines will form one continued straight line. For since CEA is equal to BED (a), and CEB is equal to AED (a), therefore CEA and CEB taken together are equal to BED and AED taken together (b); and because the whole four taken together are equal to four right angles (c), therefore CEA and CEB taken together are equal to two right angles, and consequently AE and EB form one continued straight line (d). And in like manner that CE and ED also form one continued straight line. (c) I. 13, cor. 3. (d) I. 14. it may be shown PROPOSITION XVI. THEOREM.—If one side (BC) of a triangle (ABC) be produced, the external angle (ACD) is greater than either of the internal opposite angles (ABC or A). CONSTRUCTION. Bisect AC in E (a), draw BE, and produce it until EF is equal to BE. Also join FC. DEMONSTRATION. Because in the triangles EAB and ECF the side EA is equal to the side EC (b), the side EB to the side EF (b), and the angle AEB is equal to the angle FEC (c), therefore the angle A is equal to the angle ECF (d). And therefore ACD being greater than ECF, is also greater than A. In like manner, if the side AC be produced, it may be proved that the angle BCG, and therefore its equal ACD (c), is greater than the angle ABC. B COROLLARY 1. If from any point (C) two straight lines be drawn to meet a third straight line (AB), one of them (CD) perpendicular to it, and the other (CE) not; then that which is perpendicular shall be on that side of the other on which it forms an acute angle. I. 10. (b) Constr. A E (a) Def. 10. (b) I. 16. For if it be possible, let it fall on the same side as the obtuse angle CEA; then the angle CDA, being a right angle, is less than CEA (a); but CDA is also greater than the internal opposite angle CEA (b), which is absurd; therefore CD cannot fall on the side next the obtuse angle, but on the same side as the acute angle. COROLLARY 2. If two straight lines be drawn from any point (C) to the same straight line (AB); they cannot both be perpendicular to it. For if it be possible, let both CD and CE be perpendicular to AB; then the angle ADC is equal to AEC (a); but ADC is also greater than AEC (6), which is absurd; therefore the straight lines CD and CE cannot both be perpendicular to AB. (a) Ax. 11. (b) I. 16. SCHOLIUM. Both the sixteenth and seventeenth propositions are included in the thirty-second, and as they are not employed until subsequent to that proposition, they might have been omitted in this place. PROPOSITION XVII. THEOREM.-If any two angles are those of a triangle (ABC), they are together less than two right angles. Then D DEMONSTRATION. Produce BC to D. the external angle ACD is greater than the internal opposite angle B (a); and if to both the angle ACB be added, ACD and ACB taken together are greater than B and ACB taken together; but ACD and ACB taken together are equal to two right angles (b), therefore B and ACB, two of the angles of the triangle ABC, are together less than two right angles. And in like manner it may be proved that either the angles A and B or A and ACB are together less than two right angles. PROPOSITION XVIII. (a) I. 16. (b) I. 13. THEOREM.-If one side (AC) of any triangle (ABC) be greater than another (AB), the angle (ABC) opposite to the greater side is greater than the angle (C) which is opposite to the less. CONSTRUCTION. From the greater side AC cut off AD equal to the less AB (a), and join BD. I. 3. Constr. DEMONSTRATION. Because in the triangle ABD the sides AD and AB are equal (b), therefore the angles ADB and ABD are equal (c). But the angle ADB, being the external angle of the triangle BCD, is greater than the internal opposite angle C (d), therefore the angle ABD is greater than the angle C; and as ABC is greater than ABD, therefore ABC, the angle opposite the greater side, is greater than C, the angle opposite the less. PROPOSITION XIX. (c) I. 5. (d) I. 16. THEOREM.-If in any triangle (ABC) one angle (B) is greater than another (C), the side (AC) which is opposite to the greater angle is greater than the side (AB) which is opposite to the less. DEMONSTRATION. For the side AC must either be equal to, less than, or greater than AB. It is not equal to AB, because then the angle B would be equal to the angle C (a), while by the B (a) I. 5. hypothesis, B is greater than C. And it is not less than AB, because then the angle B would be less than C (b); therefore the side AC opposite to the greater angle is greater than the side AB opposite to the less. SCHOLIA. 1. This proposition is the converse of the preceding proposition, and bears the same relation to proposition vi. that the preceding proposition does to proposition v. This relation is seen by combining them in the following manner. One angle of a triangle is greater or less than another (prop. xviii.), or equal to it (prop. v.), according as the side opposed to the one is greater than, less than, or equal to the side opposed to the other. And one side of a triangle is greater or less than another (prop. xix.), or equal to it (prop. vi.), according as the angle opposed to the one is greater than, less than, or equal to the angle opposed to the other. The mutual relation of these four propositions may also be shown in the following manner: PROP. V.-If ABAC, then C▲ B. PROP. XVIII.-Iƒ AB> AC, then C> ▲ B. The propositions connected by a bracket are the converse of each other, because that which is the hypothesis in the one is the predicate in the other. 2. The form of argument employed in this proposition is the "reductio ad absurdum." PROPOSITION XX. THEOREM.-Any two sides (BA and AC) of a triangle (ABC) are together greater than the third side (BC). CONSTRUCTION. Produce either of the two sides, as BA, until AD is equal to AC, the other side, and join CD. (a) Constr. DEMONSTRATION. Because the triangle ACD is isosceles (a), the angles ACD and D are equal (6). But the angle BCD is greater than ACD (c), therefore it is greater than the angle D. Now because in the triangle BCD the angle BCD is greater than the angle D, therefore the side BD opposite BCD is greater than the side BC opposite D (d). BA and AC (a), therefore the sides BA and AC are together greater than the third side BC. But BD is equal to COROLLARY. It follows from this proposition that the difference between any two sides of a triangle is less than the third side; for since BA and AC are together greater than BC, if AC be subtracted from each, BA will be greater than the difference between BC and AC. |