equal to DE, and BC to EF, the two fides GB, BC, are equal Book 1. to the two DE, EF, each to each ; and the angle GBC is equal \le^-) to the angle DEF; therefore the base GC is equal" to the base a 4.1. DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal fides are opposite ; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothesis, equal to the angle BCA ; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; the base therefore AC is equal" to the base DF, and the third angle BAC to the third angle EDF. AC to DF, and BC For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal fides are opposite; therefore the angle BHA is equal to the angle EFD ; but EFD is equal to the angle BCA ; therefore also 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 impossible b5 wherefore BC is not unequal to EF, that is, b 14-1. it is equal to it; and AB is equal to DE; therefore the two AB, BC are equal to the two DE, EP, 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. 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 interior and opposite angle EFG 3 but it is also equal to it, which is impossible; therefore AB and CD being produced do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards A, C ; but those straight lines which meet neither way, though produced ever so far, are parallel b to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. angle EGB equal" to the angle AGH, the angle AGH is equal Book I. to the angle GHD; and they are the alternate angles; therefore Q_v^y AB is parallel b to CD. Again, because the angles BGH, GHD a 1s. 1. are equal * to two right angles, and that AGH, BGH are also ! ##, equal “to two right angles; the angles AGH, BGH are equal i ; op. 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. a straight fire falls upon two parallel straight lines, it i. makes the alternate angles equal to one another; and jo. the exterior angle equal to the interior and opposite upon ition. the same fide; and likwise the two interior angles upon the same side together equal to two right angles. a 29. 1. b 27, 1. a 23. I TRA1GHT lines which are parallel to the same straight line are parallel to one another. Let A be the given point, and BC the given straight line; it is required to draw a straight line E A F through the point A, parallel to the straight line BC. In BC take any point D, and join AD; and at the point A in the straight line AD make the angle B D C 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 o EAF is drawn through the given point A parallel to the given Book I. flight line BC. Which was to be done. \-en-V b 27, 1. EA F a fide 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 fides 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 are together equal to two right angles. Through the point C draw CE parallel " to the straight a 31. 1. line AB; and because AB is A parallel to CE and AC meets IE them, the alternate angles BAC, ACE are equal". A- b 29. 1. gain, because AB is parallel to CE, and BD falls upon – them, the exterior angle ECD B C ID 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 equals 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 right angles; therefore also the c 13. 1. angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a fide of a triangle, &c. Q. E. D. CoR. 1. All the interior angles of any rectilineal figure, together D with four right angles, are equal to twice as many right angles as E the figure has fides. C For any rectilineal figure ABCDE can be divided into as many triangles as the figure has fides, by drawing straight lines from a point F within the figure A. B |