line EAF is drawn through the given point A parallel to the Book 1. given straight line BC. Which was to be done. PROP. XXXII. THEOR. F a fide of any triangle be produced, the exterior 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 CE parallel to the ftraight 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. t 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 thefe angles add the angle ABC, 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. 2. right angles; therefore also the angles CBA, BAC, ACB aré equal to two right angles. Wherefore, if a fide of a triangle, &c. Q. E. D. COR. I. All the interior angles of any rectilineal figure, are equal to twice as many right angles as the figure has fides, wanting four right angles. E 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 all to each of its angles. And, by the preceding propofition, D 34 Book I. all the angles of these trianges are equal to twice as many a 2. Cor. 15. 1. b 13. I. * 29. I. gles, that is, as there are fides of Ses are equal to the angles of angles at the point F, which is ngles: that is, together with twice as many right angles as all the angles of the figure, that is, the angles of the fiany right angles as the figure right angles as there are t COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every in terior angle ABC, to twice as many right D the figure, are equal angles as there are fides of the figure; 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 exterior angles are equal to four right angles. THE PROP. XXXIH. THEOR. HE ftraight lines which join the extremities of two equal and parallel ftraight lines, towards the fame parts, are alfo themfelves equal and parallel. Let AB, CD be equal and A parallel straight lines, and joined towards the fame parts by the ftraight lines AC, BD; AC, BD are also equal and parallel. Join BC; and because AB is parallel to CD, and BC meets D them, the alternate angles ABC, BCD are equal; and be caufe caufe AB is equal to CD, and BC common to the two tri- Book I. angles ABC, DCB, the two fides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal b to the base BD, and the tri- b 4. 8. angle 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 straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was fhewn to c 27. 1. be equal to it. Therefore, ftraight lines, &c. Q. E. D. TH HE oppofite fides and angles of a parallelogram are equal to one another, and the diameter bifects it, that is, divides it in two equal parts. N. B. A Parallelogram is a four-fided figure, of which the oppofite fides are parallel; and the diameter is the ftraight 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 angles ABC, BCD are equal to one another; and becaufe AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal a C B a 29. i. D to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles t BCD, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal D 2 angles; 1 26. I. Book I angles; therefore their other fides are equal, each to each, and the third angle of the one to the third angle of the other, viz. the fide AB to the fide CD, and AC to BD, and the angle BAC equal to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle AB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been hown to be equal to the angle BDC; therefore the oppofite fides and angles of a parallelogram are equal to one another: alfo, its diameter bifects it, 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. Therefore, &c. Q. E. D. 4. I. See the 2d and 3d figures. ARALLELOGRAMS upon the fame bafe and between the fame parallels, are equal to one a nother. Let the parallelograms ABCD, EBCF be upon the fame bafe BC, and between the fame parallels AF, BC; the parallelogram ABCD is equal to the parallelogram EBCF. But, if the fides AD, EF, oppofite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the fame point; then, because ABCD is a parallelogram, AD is equal a to BC; for the fame reason EF is equal to BC; wherefore AD is equal to EF; and DE is common; therefore 2. or 3. the whole, or the remainder, AE is equal to the whole, or 6 r. Ax. Ax. C the the remainder DF; now AB is also equal to DC; therefore Book I. the two DE FA ED F WW B B e 4. r. EA, AB are equal to the two FD, DC, each to each; but the exterior angle FDC is equal to the interior EAB, d 19.1. wherefore the bafe EB is equal to the base FC, and the triangle EAB to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB; the remainders will then be equalf, that is, the parallelogram ABCD is equal to the f 3. Ax. parallelogram EBCF. Therefore, parallelograms upon the fame bafe, &c. Q. E. D. P ther. ARALLELOGRAMS upon equal bases, and be- a F G because BC is equal to FG, and FG to EH, BC is equal to a 34. 1. EH; and they are parallels, and joined 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; therefore EB, CH are b 33. x. both equal and parallel, and EBCH is a parallelogram; and C it is equal to ABCD, because it is upon the fame base BC, c 35. 1. D 3 and |