Book. I. a 15. 1. b 27. 1. I PROP. XXVIII. THEOR. F a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be 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 oppofite angle GHD upon the same fide; E G B H D F Because the angle EGB C is equal to the angle GHD, and the angle EGB equal a to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel b to c By Hyp. CD. Again, because the angles BGH, GHD are equal e to two right angles; and that AGH, BGH, are alfo equald 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. d 13. 1. See N. PROP. XXIX. THEOR. Fa straight line fall upon two parallel straight it makes the alternate angles equal to one another; and the exterior angle equal to the interior and oppofite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles. Let Let the ftraight line EF fall upon the parallel straight lines Book I. AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and oppofite, upon the fame fide, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL another, which is impossible b. The angles E brr. Ax. C 15. 1. AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGHc; and AGH is proved to be equal to GHD; therefore EGB is likewife 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 equald to d 13. 1. two right angles; therefore alfo 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 GH will meet on the fide 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 fide of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet toward the points L and D; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles d, 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, fince KL and CD are not parallel, and do not meet towards L. and D, they will meet if produced towards K and C. PROP. 32 Book I. S PROP. XXX. THEOR. TRAIGHT lines which are parallel to the fame straight line are parallel to one another. Let AB, CD be each of them parallel to EF; AB is alfo parallel to CD. Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle * 29. 1. AGH is equala to the angle GHF. Again, because the A CD, the angle GHF is equal a to the angle GKD; and it K was shewn that the 'angle c D AGK is equal to the angle GHF; therefore alfo AGK is equal to GKD; and they are alternate angles; therefore b 27. F. AB is parallel b to CD. Wherefore straight lines, &c. Q E. D. T PROP. XXXI. PRO B. O 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 A F C a 23. I. D, and join AD; and at Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal b 27. 1. to one another, EF is parallelb 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. PROP. XXXII. THEOR. IF a fide of any triangle be produced, the exterior angle is equal to the two interior and oppofite 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 a to the straight line AB; and because AB is pa rallel to CE and AC meets them, the al ternate angles BAC, B ACE are equalb. 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 oppofite 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 c131. right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a fide of a triangle, &c. Q. E. D. many triangles as the figure has fides, by drawing straight lines from a point F within the figure A B to each of its angles. And, by the preceding propofition, D all Book I. all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are fides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is a 2. Cor. the common vertex of the triangles: that is a, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides. 15. 1. b 13. 1. COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every in terior angle ABC, A that is by the forego ing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the ex terior angles are equal to four right angles. PROP. XXXIII. THEOR. THE ftraight lines which join the extremities of two equal and parallel straight lines, towards the fame parts, are also themselves equal and parallel. Let AB, CD be equal and A parallel straight lines, and joined towards the same parts by the straight lines A C, BD; AC, BD are also equal and parallel. B Join BC; and because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equala; and be a 29. 1. oaufe |