Let the straight line EF fall upon the parallel straight lines AB, CD. Then the alternate angles AGH, GHD shall be equal to one another; the exterior angle EGB shall be equal to the interior and opposite angle GHD upon the same side of the line EF; and the two interior angles BGH, GHD upon the same side of EF shall be together equal to two right angles. C E H Demonstration. First. For, if the angle AGH be not equal to the alternate angle GHD, one of them must be greater than the other; if possible, let AGH be greater than GHD, then because the angle AGH is greater than the angle GHD, add to each of these unequals the angle BGH, therefore B 1. The angles AGH, BGH are greater than the angles BGH, GHD (Ax. 4), but the angles AGH, BGH are equal to two right angles (I. 13), therefore 2. The angles BGH, GHD are less than two right angles; but those straight lines, which with another straight line falling upon them, make the two interior angles on the same side less than two right angles, will meet together if continually produced (Ax. 12); therefore • 3. The straight lines_AB, CD if produced far enough will meet towards B, D ; but they never meet, since they are parallel by the hypothesis ; therefore on the same side of the line. 4. The angle AGH is not unequal to the angle GHD, that is, the angle AGH is equal to the alternate angle GHD. Secondly. Because the angle AGH is equal to the angle EGB (I. 15), and the angle AGH is equal to the angle GHD, therefore 5. The exterior angle EGB is equal to the interior and opposite angle GHD, Thirdly. Because the angle EGB is equal to the angle GHD, add to each of them the angle BGH ; therefore 6. The angles EGB, BGH are equal to the angles BGH, GHD (Ax. 2); but EGB, BGH are equal to two right angles (I. 13), therefore also 7. BGH, GHD are equal to two right angles (Ax. 1). Wherefore, if a straight line, &c. Q.E.D. PROPOSITION 30.—Theorem. Straight lines which are parallel to the same straight line are parallel to each other. Let the straight lines AB, CD, be each of them parallel to EF. Then shall AB be also parallel to CD. Construction. Let the straight line GHK cut AB, EF, CD. Demonstration. Then because GHK cuts the parallel straight lines AB, EF, in G, H, therefore 1. The angle AGH is equal to the alternate angle GHF (I. 29). Again, because GHK cuts the parallel straight lines EF, CD in H, K, therefore 3. The angle AGH is equal to the angle GKD, 2. The exterior angle GHF is equal to the interior angle HKD, and it was shown that the angle AGH is equal to the angle GHF; therefore and these are alternate angles; therefore 4. AB is parallel to CD (I. 27). Wherefore, straight lines which are parallel, &c. Q.E.D. PROPOSITION 31.-Problem. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw, through the point A, a straight line parallel to the straight line BC. E B F D Construction. In the line BC take any point D, and join AD; at the point A in the straight line AD, make the angle DAE equal to the angle ADC (I. 23) on the opposite side of AD; and produce the straight line EA to F. Then EF shall be parallel to BC. Proof. Because the straight line AD meets the two straight lines EF, BC, and makes the alternate angles EAD, ADC equal to one another, therefore EF is parallel to BC (I. 27). Wherefore, through the given point A, a straight line EAF has been drawn 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 together equal to two right angles. E B Let ABC be a triangle, and let one of its sides BC be produced to D. Then the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC: and the three interior angles ABC, BCA, CAB shall be equal to two right angles. Construction. Through the point C draw CE parallel to the side BA (I. 31). Demonstration. Then, because CE is parallel to BA, and AC meets them, therefore 1. The angle ACE is equal to the alternate angle BAC (I. 29). Again, because CE is parallel to AB, and BD falls upon them, therefore 2. The exterior angle ECD is equal to the interior and opposite angle ABC (I. 29); but the angle ACE was shown to be equal to the angle BAC, therefore 3. The whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC (Ax. 2). Again, because the angle ACD is equal to the two angles ABC, BAC, to each of these equals add the angle ACB, therefore 4. The angles ACD and ACB are equal to the three angles ABC, BAG, and ACB (Ax. 2); but the angles ACD, ACB are equal to two right angles (I. 13), therefore also 5. The angles ABC, BAC, ACB are equal to two right angles (Ax. 1). Wherefore, if a side of any triangle be produced, &c. Q.E.D. Cor. 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. D B Demonstration. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point within the figure to each of its angles. Then, because the three interior angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides, E D therefore all the angles of these triangles are equal to twice as many right angles as the figure has sides; B but the same angles of these triangles are equal to the interior angles of the figure together with the angles at the point F; 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); therefore the same angles of these triangles are equal to the angles of the figure together with four right angles; D but it has been proved that the angles of the triangles are equal to twice as many right angles as the figure has sides; 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. Cor. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. B C Demonstration. Since every interior angle ABC, with its adjacent exterior angle ABD, is equal to two right angles (I. 13), therefore all the interior angles, together with all the exterior angles, are equal to twice as many right angles as the figure has sides; but it has been proved by the foregoing corollary, that all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides; |