EFD; but EFD is equal to the angle BCA; therefore also Book I. the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA, which is impossibleb; wherefore b 16. 1. BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D. PROP. XXVII. THEOR. IF a straight line falling upon two other straight lines make the alternate angles equal to one another, these two straight lines are parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another: AB is parallel to CD. For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C ; let them be produced and meet towards B, D, in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater than the a 16.1. interior and opposite angle EFG; but it is also equal to it, which A is impossible; therefore AB and CD being pro G D duced, do not meet towards B, D. In like manner it may be de c monstrated that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel to one another. AB therefore is parallel b 30. Def. to CD. Wherefore, if a straight line, &c. Q. E. D. Book I. " a 15. 1. b 27. 1. d 13. 1. dł PROP. XXVIII. THEOR. IF a straight line falling upon two other straight lines make the exterior angle equal to the interior and opposite upon the same side of the line, or make the interior angles upon the same side together equal to two right angles; the two straight lines are parallel to one another. Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; Because the angle EGB E angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel to CD. c by Hyp. Again, because the angles BGH, GHD are equal to two right angles; and AGH, BGH are also equal to two right angles; the angles AGH, BGH are equal to the angles BGH, GHD: take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. See N. PROP. XXIX. THEOR. 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 likewise the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite angle GHD, upon the same side; and the two interior angles BGH, GHD, upon the same side, are together equal to two right angles. Book I. For if AGH be not equal to GHD, let KG be drawn, making the angle KGH equal to GHD, and produce a 23. 1. KG to L; then KL will be parallel to E possible. The angles c 11. Ax. AGH, GHD therefore are not unequal; that is, they are equal to one another. Now, the angle EGB is equal to AGH; and AGH is proved to be equal to GHD; there- d 15. 1. fore EGB is likewise equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to e 13. 1. two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D. COR. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles. For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet on the other side of EF; for the angles LGH, GHD would then be two angles of a triangle, and less than two right anglesf; but this is impossible; for the f 17. 1. four angles KGH, HGL, CHG, GHD are together equal to four right angles, of which the two KGH, CHG are by supposition less than two right angles; therefore the other two HGL, GHD are greater than two right angles. Therefore, since KL and CD are not parallel, and do not meet towards L and D, they will meet if produced towards K and C. Book I. a 29. 1. PROP. XXX. THEOR. STRAIGHT lines which are parallel to the same straight line are parallel to one another. Let AB, CD be each of them parallel to EF; AB is also parallel to CD. Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to the angle A GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal a to the angle GKD; and it was shown that the angle C AGK is equal to the angle GHF; therefore also AGK E G B H F K D is equal to GKD; and they are alternate angles; therefore AB is parallel to CD. b 27. 1. Q. E. D. a 23. 1. b 27. 1. Wherefore, straight lines, &c. PROP. XXXI. PROB. 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 a straight line through the E point A, parallel to the straight line BC. In BC take any point B D A ; F C the point A, in the straight line AD, makea the angle DAE equal to the angle ADC; and produce the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. Therefore the straight line EAF is drawn through the given point A parallel to the Book I. given straight line BC. Which was to be done. 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. Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB, ar etogether equal to two right angles. Through the point C draw CE parallela to the straight line AB; and because AB is parallel to CE, and AC meets them, the al ternate angles BAC, B ACE are equal. Again, because AB is parallel to CE, and b 29. 1. BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal to two c 13. 1. right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. fror a point F within the figure D to each of its angles. And, by the preceding proposition, E |