ACB, BAC; that is, equal to two right angles. Where-Book І. fore, &c. Cor. 1. Hence all the three angles of any one triangle are b Ax. 1. equal to all the three angles of any other triangle, either sepa. c 13. rately or taken together. 2. If two angles of one triangle be equal to two angles of another triangle, either separately or together, the remaining angle of the one is equal to the remaining angle of the o ther. 3. If one angle of a triangle be a right one, the other two angles are together equal to a right angle. 4. If the angle included by the equal fides of an isofceles triangle be a right one, each of the other angles will be half a right one. 5. Any angle in an equilateral triangle is one third of two right angles, or two thirds of one right angle. 6. If one angle of a triangle be equal to the other two, that angle is a right one; for, if the fide is produced, the adjacent angle is equal to the other two ; therefore each of them are right angles. 7. All the inward angles of any right lined figure make twice as many right angles, abating four, as the figure has fides. For any right lined figure can be divided into triangles, the inward angles of each equal to two right angles, and all the triangles together equal to the number of fides of the figure, abating two : Therefore all the inward angles will be equal to twice the number of fides, abating four. 8. All the outward angles of any right lined figure are equal to four right angles. For all the outward and inward angles together are equal to double the number of fides; but the inward angles are equal to double the number of fides, abating Therefore the outward are equal to four right angles. four: PROP. XXXIII. THEOR. F two right lines join two equal and parallel right lines toward the same part, these lines will be equal and parallel. Let AB, CD, be two equal and parallel right lines; join AC, BD; then will the right lines AC, BD, be equal and parallel. For, because AB, CD, are parallel, and BC falls upon them, the angle ABC is equal to BCD2; but, because AB is equal to a 29. CD, and BC common; and the angle ABC equal to BCD, the bafe AC is equal to BD, and the angle ACB equal to CBD b; b 4. but Book I. but these are alternate angles: Therefore AC is parallel to 1, 29. BD, and likewise equal. Wherefore, &c. PROP. XXXIV. THEOR. HE opposite fides and opposite angles of every parallelogram are equal; and the diameter divides it into two equal parts. Let ABCD be a parallelogram, the opposite fides AB, CD; AC, BD, are equal; the angle CAB equal to BDC, and ACD to ABD; and the diameter BC bisects it. 1 29. b 26. C 4. For, because AB is parallel to CD, and BC falls upon them, the angle ABC is equal to BCD. For the fame reason ACB is equal to CBD; therefore the two angles ABC, ACB, in the triangle ABC, are equal to the two angles CBD, BCD, in the triangle BCD; and the fide BC common to both: Therefore the two fides AC, AB, of the one triangle, are equal to the two fides BD, DC, of the other, each to each; and the angle BAC equal to BDC, and ACD to ABDb. Again, because the two fides AC, AB, are equal to the two fides BD, DC, each to each, and the angle CAB equal to BDC, the bafe BC common: Therefore the triangles are equal; and BC bisects the parallelogram. Wherefore, &c. a 34. b Ax. 1. с Ах. 6. d Ax. 2. € 34. f 29. g 4. h Ax. 3. PROP. XXXV. and XXXVII. THEOR. D Arallelograms and triangles, conftitute upon the fame bafe, jelves, viz. parallelogram to parallelogram, and triangle to triangle. Let ABCD, EBCF, be two parallelograms (Fig. 2.) constitute upon the fame base BC, and between the fame parallels BC, AF; the parallelograms ABCD, EBCF, are equal. For, because AD, EF, are each equal to BC, they are equal to one another b. If the point E coincide with D. (Fig. 1.) each of the parallelograms are double the triangle DBC; therefore equal to one another. If AD is less than AE, add DE to both; then the whole AE is equal to DF d, DC to AB, and the angle FDC to EABf: Therefore the triangles FDC, EAB, are equal 8. Take DGE from both; the trapeziums, ADGB, FEGC, are equalh. Add the triangle GBC to both; then the whole parallelogram ABCD is equal to the parallelogram EBCF d. If AD is greater than AE, take DE from both; then Booк І. the remainder AE will be equal to DFh, the triangle AEB to ~ DFC. Add EBCD to both, then the parallelograms ABCD, d Ax. 2. EBCF, are equald: So, likewise, if the diameters AC, BF, beh Ax. 3. drawn, then the triangle ABC will be equal to FBC. Where- i 34. and fore, &c. Cor. Hence every parallelogram is equal to a right angled parallelogram, constitute upon the fame base, and between the fame parallels; and every triangle constitute upon the same base, and betwixt the fame parallels, is half the rectangle. P PROP. XXXVI. and XXXVIII. THEOR. Arallelograms and triangles, constitute upon equal bafes, and parallelogram to parallelogram, and triangle to triangle. ت Let the parallelograms ABCD, EFGH, be constitute upon the equal bases BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD will be equal to EFGH. Ax. 7. For, join EB, CH, the parallelograms AC, EG *, are each equal to the parallelogram EC2; therefore equal to one ano-a 35. ther b. Join AC, FH; then the triangles ABC, HGF, are e-b Ax. r. qual. Wherefore, &c. PROP. XXXIX. THEOR. QU AL triangles, constitute upon the fame bafe, on the fame Let the equal triangles ABC, DBC, be conftitute upon the fame base, BC, on the same side; the right line AD, that joins their vertex, will be parallel to BC. c 34. and Ax. 7. If not, draw AE, parallel to BC; join EC; then the triangles ABC, EBC, are equal; but DBC is equal to ABCb; a 35. therefore DBC, EBC, are equal, a part to the whole; which b Hype is impossible. Therefore no line but AD is parallel to BC. Wherefore, &c. PROP. * Parallelograms arc expressed by the letters at the opposite angles. Воок І. a 36. b hyp. a 35. 34. a 10. b 23. C 31. d 36. e 41. fAx. 6. PROP. XL. THEOR. QUAL triangles, constitute upon equal bases, on the fame fide, are between the fame parallels. Let ABC, DGE, be equal triangles, constitute upon the equal bases BC, GE, on the same fide; then AD is parallel to BE. If not, draw AF parallel to BE; join FE; then the triangle ABC is equal to FGE; but DGE is equal to ABC; therefore DGE is equal to FGE, a part to the whole; which is impossible: Therefore AD is parallel to BE. Wherefore, &c. PROP. XLI. THEOR. F a parallelogram and triangle be conftitute upon the same base, and between the same parallels, the parallelogram will be double the triangle. Let the parallelogram be ABCD, and triangle EBC, having the fame base BC, and be between the fame parallels AE, BC, the parallelogram ABCD is double the triangle EBC; join AC. Then the triangle ABC is equal to EBC2; but the parallelogram ABCD is double the triangle ABCb; and therefore double EBC. Wherefore, &c. PROP. XLII. PROB. O constitute a parallelogram equal to a given triangle, having an angle in it eq equal to a given right lined angle. Let the given triangle be ABC, and right lined angle D, it is required to conftitute a parallelogram equal to the given triangle ABC, having an angle in it equal to D. a Bilect BC in E2; make an angle CEF equal to Db; through A draw AG parallel to CE ; and through C draw CG parallel to EF; join AE; then the triangles ABE, AEC, are equal d; and ABC is double AEC; but the parallelogram EG is double the triangle EAC: Therefore the parallelogram EG is equal to the triangle ABCf, and the angle FEC equal to D. Wherefore, &c. I PROP. XLIII. THEOK. N every parallelogram, the complements that stand about the diameter are equal to one another. Let ABCD be a parallelogram; BD its diameter; the parts of which BK, KD, the diameters of the parallelograms HKFD, EBGK; the remaining parallelograms ΑΕΚΗ, KGCF, its complements, are equal to one another. For, because DB is the diameter of the parallelogram ABCD, the triangles ADB, DBC, are equal. For the fame reason, a 34 the triangle HKD is equal to DFK, and EBK to BKG; wherefore the triangles HKD and EKB are equal to DFK, BKG, b. bAx. 2. Take HKD, EKB, from ADB, and DFK, BKG, from DBC, c Ax 3. there remains ΛΕΚΗ equal to KGCF. Wherefore, &c. angle. PROP. XLIV. PROB. O apply a parallelogram to a given right line equal to a given It is required, upon the given right line AB, to make a parallelogram equal to a given triangle C, having an angle in it equal to a given angle D. Make the parallelogram FGBE equal to the triangle C, having the angle EBG equal to D2; put BE in a right line with a 42. AB; and produce FG to H; through A draw AH parallel to GB, or FE; join HB. Now, because the angles EFH, FHA, are equal to two right angles, the angles EFH, FHB, are less b 29. than two right angles; then FE, HB, being produced, will meet in some point; which let be K; through which draw KL parallel to FH; and produce GB, HA, to M, L ; wherefore FHLK is a parallelogram, whose diameter is HK; and whofse complements FGBE, BALM, are equald; but FGBE, was d 43. made equal to C; and the angle EBG equal to D; therefore BALM is equal to C, and the angle ABM equal to D. c 15. Wherefore, &c. PROP. XLV. PR B. O make a parallelogram equal to a given right lined figure, C 17, Core |