5. Therefore the remaining angle AGH is equal to the remaining angle GHD (ax. 3), and they are alternate angles. 6. Therefore AB is parallel to CD. (I. 27.) Conclusion. Therefore, if a straight line, &c. Q. E. D. PROPOSITION 29.-THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite upon the same side; and also the two interior angles upon the same side together equal to tro right angles. (References—Prop. I. 13, 15; ax. 1, 2, 4, 12.) Hypothesis. Let the straight line EF fall upon the parallel straight lines AB, CD. Sequence.-1. The alternate angles AGH, GHD, shall be equal to one another. 2. The exterior angle EGB shall be equal to GHD, the interior and opposite angle upon the same side.. 3. And the two interior angles on A -B the same side, BGH, GHD, shall be together equal to two right angles. Hypothesis.-(II.) For if AGH be not equal to GHD, one of them must be greater than the other. Let AGH be the greater. Demonstration.-l. Then the angle AGH is assumed to be greater than the angle GHD, to each of them add the angle BGH. 2. Therefore the angles AGH, BGH, are greater than the angles BGH, GHD. (ax. 4.) 3. But the angles ÀGH, BGH, are together equal to two right angles. (I. 13.) 4. Therefore the angles BGH, GHD, are less than two right angles. 5. But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together, less than two right angles, these straight lines being continually produced, shall at length meet on that side on which are the angles which are less than two right angles. (ax. 12.) 6. Therefore the straight lines AB, CD, will meet if pro duced far enough. 7. But they cannot meet, because they are parallel straight lines. (hyp.) 8. Therefore the angle AGH is not unequal to the angle GHD; that is, it is equal to it. 9. But the angle AGH is equal to the angle EGB. (I. 15.) 10. Therefore the angle EGB is equal to the angle GHD. (ax. 1.) 11. Add to each of these the angle BGH. 12. Therefore the angles EGB, BGH, are equal to the angles BGH, GHD. (ax. 2.) 13. But the angles ÉGB, BGH, are equal to two right angles. (I. 13.) 14. Therefore also BGH, GHD, are equal to two right angles (ax. 1.) Conclusion. Therefore, if a straight line, &c. Q.E.D. PROPOSITION 30.-THEOREM. Straight lines which are parallel to the same straight line are parallel to one another. A B (References-- Prop. I. 27, 29; ax. 1.) Hypothesis. Let AB, CD, be each of them parallel to EF. Sequence.-AB shall be parallel to CD. Construction.—Let the straight line GHK cut AB, EF, CD. Demonstration. . 1. Because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to the angle GHF. (I. 29.) 2. Again, because GK cuts the H parallel straight lines EF, CD, the angle GHF is equal to the angle cGÁD. (I. 29.) 3. And it was shown that the angle AGK is equal to the angle GHF. 4. Therefore the angle AGK is equal to the angle GKD (ax. 1), and they are alternate angles. 6. Therefore AB is parallel to ČD. (T.. 27.) Conclusion. Therefore, straight lines, &c. Q. E. D. E A F PROPOSITION 31.-PROBLEM. To draw a straight line through a given point, parallel to a given straight line. (References-Prop. I. 23, 27.) Given.—Let A be the given point, and BC the given straight line. Sought. It is required to draw a straight line through the point A, parallel to BC. Construction.-1. In BC take any point D, and join AD. 2. At the point A, in the straight line AD, make the angle DAE equal to the angle ADC. (I. 23.) B 3. Produce the straight line EA to F. Then EF shall be parallel to BC. Proof.-1. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate augles EAD, ADC, equal to one another. 2. Therefore EF is parallel to BC. (I. 27.) Conclusion.-Therefore, the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Q. E. F. PROPOSITION 32.—THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. (References-Prop. I. 13, 29, 31; ax. 1, 2.) Hypothesis.—Let ABC be a triangle, and let one of its sides BC be produced to D. Sequence.-1. The exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC. 2. And the three interior angles of the triangle, namely, ABC, BCA, CAB, shall be equal to two right angles. Construction.-Through the point C, draw CE parallel to AB. (1. 31.) Demonstration.–1. Because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE, are equal . (I. 29.) B 2. Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. (1. 29.) 3. But the angle ACE was shown to be equal to the angle BAC. 4. Therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, A BC. (ax. 2.) 5. To each of these equals add the angle ACB. 6. Therefore the angles ACD, ACB, are equal to the three angles CBA, BAC, ACB. (ax. 2.) 7. But the angles ACD, ACB, are equal to two right angles (I. 13.) 8. Therefore also the angles CBA, BAC, ACB, are equal to two right angles. (ax. 1.) Conclusion. Therefore, if a side of any triangle, &c. Q. E.D. Corollary:-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. 1. For any rectilineal figure ABCDE can, by drawing straight lines from a point F within the figure to each angle, be divided into as many triangles as the figure has sides. 2. And, by the preceding proposition, the angles of each triangle are equal to two right angles. 3. Therefore all the angles of the triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure. 4. But the same angles are equal to the angles of the figure, together with the angles at the point F. 5. 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.) 6. 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. Corollary.-2. All the exterior angles of any rectilineal figure are together equal to four right angles. 1. The interior angle ABC, with its adjacent exterior angle ABD, are equal to two right angles. (I. 13.) 2. Therefore all the interior, together with all the exterior / angles of the figure, are equal to twice as many right angles as the figure has sides. 3. But all the interior angles, together with four right angles, are equal to twice as many right angles А 4. Therefore all the interior 5. Take away the interior angles which are common. 6. Therefore all the exterior angles are equal to four right angles. (ax. 3.) B 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. (References-Prop. I. 4, 27, 29.) Sequence.—AC and BD shall be equal and parallel. Demonstration.-1. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal. (I. 29.) 2. 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 sides DC, CB, each to each; 3. And the angle ABC was proved to be equal to the angle BCD; 4. Therefore the base AC is equal to the base BD, (I. 4.). 5. And the triangle ABC is equal to the triangle BCD, (I. 4.) 6. And the other angles are equal to the other angles, each to each, to which the equal sides are opposite; 7. Therefore the angle ACB is equal to the angle CBD. 8. And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another; B |