Particular Enunciation. Given. The straight line EF falling on the straight lines AB, CD; and making the angle A EF equal to the alternate angle E FD. B Fig. 1 D Required. To prove that— A B is parallel to CD. Construction. Let us suppose that AB and CD are not parallel. Then if they are produced they will meet, either towards A C or B D. Let them be produced, and let us suppose that they meet at a point G (Fig. 2). Proof. Then GEF is a triangle, and, by Euc. I. 16 (a), its exterior angle But the angle A EF was given equal to the angle angle A E F is both equal to and greater than angle E FG: which is impossible. So also is the supposition impossible on which this absurdity depends, viz., that AB and CD meet if produced towards B D. In the same way we may show that A B and CD do not meet towards AC. .. They are parallel (by the definition). Q. E. D. THEOREM (Euclid I. 28). Repeat. The Enunciations of Euc. I. 13; I. 15; I. 27; and Axioms 1, 2, 3a, and 10. General Enunciation. If a straight line falling on two other straight lines, make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the interior angles on the same side together equal to two right angles, the two straight lines shall be parallel to one another. (This is really two separate theorems, and we will work them as such.) Particular Enunciation (PART I). Given. The straight line EF falling on the lines AB, CD, (a) making the exterior angle E G B equal to the interior and opposite angle G HD. Required. To prove that AB is parallel to CD. Proof. The angle E G B is equal to the angle GHD (a) and by Euc. I. 15. The angle E G B is equal to the angle A G H. .. the angle A G H is equal to the angle GHD (Axiom 1). A B F B But A G H and GHD are alternate angles. .. by Euc. I. 27, A B is parallel to C D. Particular Enunciation (PART II). Q. E. D. Given. The straight line E F falling on the lines AB, CD (a) making the angles BGH, GHD together equal to two right angles. Required. To prove that A B is parallel to CD. Proof. angles BGH, GHD are together equal to two right angles (a), and, by Euc. I. 13, angles AGH, BGH are together equal to two right angles. .. by Axioms 2 and 10, (b) angles BGH, G H D are equal to angles AGH, BG H. Take away the common angle BGH. (c) the remaining angle AGH is equal to the remaining angle GHD. but, angles A G H, GHD are alternate angles. .. by Euc. I. 27, A B is parallel to C Ď. THEOREM (Euclid I. 29). Repeat.-The enunciations of Euc. I. 13 and I. 15, and Axioms 1, 2a, 4a, and 12. General Enunciation. If a straight line fall on two parallel straight lines, it makes (a) the alternate angles equal to one another, (b) the exterior angle equal to the interior and opposite angle on the same side, (c) the two interior angles on the same side together equal to two right angles. Particular Enunciation. Given. The straight line EF, falling on the two parallel lines AB, CD. Required. To prove that (a) angle A G H is equal to the alternate angle GHD. (b) exterior angle EGB is equal to the interior and opposite angle GHD. (c) the two interior angles BGH, GHD are equal to two right angles. Proof (of a). Let us suppose that the angle AGH is not equal to the angle GHD; Then one must be greater than the other. Let us suppose that (d) angle AGH is greater than angle GHD. To each angle add the angle BGH. Then, by Axiom 4a, (e) angles AGH, BGH are greater than angles BGH, GHD. But, by Euc. I. 13, angles AGH, B G H are equal to two right angles. (f).. angles BGH, GHD are less than two right angles (e). Hence, by supposing that angle AGH is not equal to angle GHD, we are able to prove that the line E F, falling on A B and C D, makes the two interior angles BG H, G H D less than two right angles (ƒ). by Axiom 12, (g) the lines AB, CD will meet if produced. But the lines A B, CD are parallel; and parallel lines never meet. So our supposition lands us in the absurdity of proving that two lines meet, and at the same time never meet. .. Our supposition is absurd, and (h) A G H is equal to G H D. Q. E. D. |