For, if AGH be not equal to GHD, one of them must be great er than the other; let AGH be the greater; and because the angle AGH is greater than the angle GHD, add to each of them the angle BGH; therefore the angles * notes on AGH, BGH are greater than the angles BGH, GHD: but the angles AGH, BGH are equal a to two right angles; therefore a 13. e the angles BGH, GHD are lefs than two right angles: but thofe itraight lines which, with another ftraight line falling upon them, make the interior angles on the fame fide lefs than two right angles, do meet together, if continually produced; there- * 12. Ax. fore the ftraight lines AB, CD, if produced far enough, fhall See the meet: but they never meet, fince they are parallel by the hypo- this prothefis; therefore the angle AGH is not unequal to the angle pofition. GHD, that is, it is equal to it; but the angle AGH is equal b b 15. 1. to the angle EGB; therefore likewife EGB is equal to GHD; add to each of thefe the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD: but EGB, BGH are equal to two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a ftraight line, &c. QE. D. STR C PROP. XXX. THEOR. TRAIGHT lines which are parallel to the fame Let AB, CD be each of them parallel to EF; AB is also parallel to CD. Let the ftraight line GHK cut AB, EF, CD; and because GHK cuts the parallel ftraight lines AB, EF, the angle AGH is equal a to the angle GHF. Again, A because the ftraight line GK cuts the parallel ftraight lines EF, CD, the E angle GHF is equal a to the angle GKD and it was fhewn, that the : F angle AGK is equal to the angle C GHF; therefore alfo AGK is equal C 13. 4. G H K D to GKD; and they are alternate angles; therefore AB is parallel b to CD. Wherefore, ftraight lines, &c. Q. E. D. T PROP. XXXI. PROB. O draw a ftraight line through a given point pa- b 27. I. BOOK I. Let A be the given point, and BC the given straight line; it Wis required to draw a ftraight line through the point A, parallel to the ftraight line BC. In BC take any point D, and join F a 23. 1. ftraight line AD make a the angle B D b 27. 1. A F Because the ftraight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel b to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given ftraight line BC. Which was to be done. PROP. XXXII. THEOR. Fa 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 oppofite 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 a 31. 1. CE parallel a to the ftraight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are eb 29. 1. qual b. Again, because AB B C D E is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and oppofite angle ABC; but the angle ACE was fhown 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 equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, C 13. 1. BAC, ACB; but the angles ACD, ACB are equal to two right angles; therefore alfo the 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 with four right angles, are equal to twice as many right angles as the figure has fides. For there are fides of the figure; and the fame angles are equal to the angles of the figure, together with the angles at the point F, Book. I. which is the common vertex of the triangles; that is, together a 2. Cor. with four right angles. Therefore all the angles of the figure, 15. 1. together with four right angles, are equal to twice as many right angles as the figure has fides. COR. 2. All the exterior angles of any rectilineal figure, are together equal to four right angles. b Because every interior angle ABC, with its adjacent exterior ABD, is equal to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are fides of the figure; that is, by the DB foregoing corollary, they are equal C to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles, THE PROP. XXXIII. THEOR. HE ftraight lines which join the extremities of two equal and parallel ftraight lines, towards the fame parts, are alfo themselves equal and parallel. Let AB, CD be equal and pa- A. rallel ftraight lines, and joined towards the fame parts by the straight lines AC, BD; AC, BD are alfo equal and parallel. Join BC; and becaufe AB is parallel to CD, and BC meets them, C B D b 13. t. the alternate angles ABC, BCD are equal: and because AB a 29. 1. is equal to CD, and BC common to the two triangles ABC, DCB, the two fides AB, BC are equal to the two DC, CB; and the angle ABC is alfo equal to the angle BCD; therefore E 2 the b 4. I. b Book I. the bafe AC is equal to the bafe BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles each to each, to which the equal fides are oppofite; therefore the angle ACB is equal to the angle CBD; and because the ftraight line BC meets the two ftraight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is pa27. 1. rallel to BD; and it was fhown to be equal to it. Therefore ftraight lines, &c. Q. E D. TH PROP. XXXIV. THEOR. HE oppofite fides and angles of parallelograms are equal to one another, and the diameter bifects them, that is, divides them in two equal parts. N. B. Aparallelogram is a four-fided figure, of which the oppofite fides are parallel; and the diameter is the Straight line joining two of its oppofite angles. Let ACDB be a parallelogram, of which BC is a diameter; the oppofite fides and angles of the figure are equal to one another; and the diameter BC bifects it. Because AB is parallel to CD, A a and BC meets them, the alternate 29. 1. angles ABC, BCD are equal to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD a C B D are equal to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal angles; therefore their other fides fhall be equal, each to each, and the 16. 1. third angle of the one to the third angle of the other C 4. I. viz. the fide AB to the fide CD, and AC to BD, and the angle BAC equal to the angle BDC: And becaufe the angle ABC is equal to the angle BCD, and the angle CBD to the a gle ACB, the whole angle ABD is equal to the whole angie ACD: And the angle BAC has been shown to be equal to the angle BDC; therefore the oppofite fides and angles of parallelograms are equal to one another; alfo, their diameter bifects them; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. PROP, ARALLELOGRAMS upon the fame bafe, and between See N. the fame parallels, are equal to one another. Let the parallelograms ABCD, EBCF, be upon the fame bafe BC, and between the fame parallels AF, BC; the parallelogram ABCD fhall be equal to the parallelogram EBCF. Because ABCD is a parallelogram, AD is equal a to BC: a 34. I. For the fame reafon EF is equal to BC; therefore AD is equal a D E C 3. Ax. F d 29. 1. e € 4. 1. b to EF: Take each of thefe equals from the whole AF; and b 1. Ax. the remainder DF is equal to the remainder AE; DC also is equal to AB; therefore the two FD, DC are equal to the two EA, AB, each to each; and the A exterior angle FDC is equal to the interior EAB; therefore the bafe FC is equal to the base EB, and the triangle FDC equal to the triangle EAB: Take each of these equal triangles from the whole figure ABCF, and the remainders B C are equal, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Wherefore parallelograms, &c. Q E. D. P PROP. XXXVI. THEOR. ARALLELOGRAMS upon equal bafes, and between Let ABCD, EFGH be pa- A rallelograms upon equal bafes BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD is equal to EFGH. B DE H Gr there b 33. I. Join BE, CH; and becaufe BC is equal to FG, and FG to a EH, BC is equal to EH; and they are parallels, and joined a 34. I. towards the fame parts by the ftraight lines BE, CH: But ftraight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel b; fore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the 35 1. fame bafe BC, and between the fame parallels BC, AD: For the like reafon, the parallelogram EFGH is equal to the fame EBCH; therefore alfo the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. C PROP. |