9. Therefore AC is parallel to BD (I. 27); and it was shown to be equal to it.” (dem. 4.) Conclusion.—Therefore, the straight lines, &c. Q. E. D. PROPOSITION 34.—THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it, that is, divides it into two equal parts. (References-Prop. I. 26, 29; ax. 2.) B Hypothesis.—Let ACDB be a parallelogram, of which BC is a diagonal. Sequence.-1. The opposite sides and angles of the figure shall be equal to one another. 2. And the diagonal BC shall bisect it. Demonstration.—1. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal to one another. (I. 29.) 2. Because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal to one another. (I. 29.) 3. Therefore the two triangles ABC, BCD, have two angles, ABC, BCA, in the one, equal to two angles, DCB, CBD, in the other, each to each; and the side BC, adjacent to the equal angles in each, is common to both triangles. 4. Therefore the other sides are equal, each to each, and the third angle of the one to the third angle of the other; namely, AB equal to CD, AC to BD, and the angle BAC to the angle CDB. (I. 26.) 5. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; 6. Therefore the whole angle ABD is equal to the whole angle ACD. (ax. 2.) 7. And the angle BAC has been shown to be equal to the angle BDC (dem. 4). Therefore the opposite sides and angles of a parallelogram are equal to one another. 8. Also the diagonal bisects it, for the triangles ABC, BCD, are every way equal. (I. 26.) Conclusion. Therefore, the opposite sides, &c. Q. E. D. PROPOSITION 35.-THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. D B D E ܦ (References --Prop. I. 4, 29, 34; ax. 1, 2, 3, 6.) Hypothesis.—Let the parallelograms ABCD, EBCF, be on the same base BC, and between the same parallels AF, BC. Sequence. The parallelogram ABCD shall be equal to the parallelogram EBCE. Case I.-If the sides AD, DF, of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, it is plain that each of the parallelograms is double of the triangle DBC (I. 34.), and that they are therefore equal to one another. (ax. 6.) Case II.—But if the sides AD, EF, opposite to the base BC, of the parallelograms ABCD, EBCF, be not terminated in the same point ; then Demonstration. 1. Because ABCD is a parallelogram, AD is equal to BC. (1. 34.) 2. For the same reason EF is equal to BC; 3. Therefore AD is equal to EF (ax. 1), and DE is common; 4. Therefore the whole, or the remainder, AE, is equal to the whole, or the remainder, DF. (ax. 2, or 3.) 5. And AB is equal to DC, (I. 34.) 6. Therefore the two EA, AB, are equal to the two FD, DC, each to each; 7. And the exterior angle FDC is equal to the interior EAB; (I. 29.) 8. Therefore the base EB is equal to the base FC, (I. 4.) 9. And the triangle EAE equal to the triangle FDC. (I. 4.) 10. Take the triangle FDC from the trapezium ABCF, and from the same (or from a similar) trapezium ABCF, take the triangle EAB, and the remainders are equal; (ax. 3.) 11. That is, the parallelogram ABCD is equal to the paral. lelogram EBCF. Conclusion.-- Therefore, parallelograms, &c. Q. E. D. B B PROPOSITION 36.-THEOREM. D E H B Parallelograms upon equal bases, and between the same parallels, are equal to one another. (References—Prop. I. 33, 34, 35; ax. 1.) Hypothesis. Let ABCD, EFGH, be parallelograms on equal bases BC, FG, and between the same parallels AH, BG. Sequence.—The parallelogram ABCD shall be equal to the parallelogram EFGH. Construction. - Join BE, CH. Demonstration. -1. Because BC is equal to FG (hyp.), and FG to EH, (I. 34.) 2. Therefore BC is equal to EH (ax. 1); and they are parallels, and joined towards the same parts by the straight lines BE, CH. 3. But straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves equal and parallel. (I. 33.) 4. Therefore BE, CH, are both equal and parallel. 5. Therefore EBCH is a parallelogram, (def.) 6. And it is equal to the parallelogram ABCD, because they are on the same base BC, and between the same parallels BC, AH. (I. 35.) 7. For the like reason, the parallelogram EFGH is equal to the same EBCH. 8. Therefore the parallelogram ABCD is equal to the parallelogram EFGH. (ax. 1.) Conclusion.—Therefore, parallelograms, &c. Q. E. D. PROPOSITION 37.-THEOREM. Triangles upon the same base, and between the same parallels, are equal to one another. (References-Prop. I. 31, 34, 35; ax. 7.) Hypothesis.—Let the triangles ABC, DBC, be on the same base BC, and between the same parallels AD, BC. Sequence.—The triangle ABC shall be equal to the triangle DBC. A D Construction.-1. Produce AD both ways, to the points E, F. 2. Through B draw BE parallel to CA, and through C draw CF parallel to BD. (I. 31.) Demonstration.-1. Then each E of the figures EBCA, DBCF, is a parallelogram (def.), and they are equal to one another, because they are on the same base BC, and between the same parallels BC, EF. (I. 35.) 2. And the triangle ABC is half of the parallelogram EBCA, because the diagonal AB bisects it; (I. 34.) 3. And the triangle DBC is half of the parallelogram DBCF, because the diagonal DC bisects it. (I. 34.) 4. But the halves of equal things are equal; (ax. 7.) 5. Therefore the triangle ABC is equal to the triangle DBC. Conclusion.—Therefore, triangles, &c. Q. E. D. B PROPOSITION 38.-THEOREM. G, H. G D Triangles upon equal bases, and between the same parallels, are equal to one another. (References—Prop. I. 31, 34, 36; ax. 7.) Hypothesis.—Let the triangles ABC, DEF, be on equal bases BC, EF, and between the same parallels BF, AD. Sequence.—The triangle ABC shall be equal to the triangle DEF. Construction.-1. Produce AD both ways to the points 2. Through B draw BG parallel to CA, and through F draw FH parallel to ED. (I. 31). Demonstration. 1. Then each of the figures GBCA, DEFH, is a parallelogram (def.), and they are equal to one another, because they are on equal bases BC, EF, and between the same parallels BF, GH. (I. 36.) 2. And the triangle ABC is half of the parallelogram GBCA, because the diagonal AB bisects it; (I. 34.) 3. And the triangle DEF is half of the parallelogram DEFH, because the diagonal DF bisects it. (I. 34.) B с 4. But the halves of equal things are equal, (ax. 7.) 5. Therefore the triangle ABC is equal to the triangle DEF. Conclusion. Therefore, triangles, &c. Q. E. D. PROPOSITION 39.-THEOREM. A Equal triangles upon the same base, and on the same side of it, are between the same parallels. (References Prop. I. 31, 37; ax. 1.) Hypothesis.--Let the equal triangles ABC, DBC, be upon the same base BC, and on the same side of it. Sequence. They shall be between the same parallels. Construction.-1. Join AD; AD shall be parallel to BC. 2. For if it is not, through A draw AE parallel to BC (I. 31), and join EC. Demonstration.-1. The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. (I. 37.) 2. But the triangle ABC is equal to the triangle DBC. (hyp.) 3. Therefore the triangle DBC is equal to the triangle EBC (ax. 1); the greater equal to the less, which is impossible. 4. Therefore AE is not parallel to BC. 5. In the same manner, it can be demonstrated that no line passing through A can be parallel to BC, except AD; 6. Therefore AD is parallel to BC. B PROPOSITION 40.-THEOREM. Equal triangles upon the same side of equal bases, that are in the same straight line, are between the same parallels. (References-Prop. I. 31, 38; ax. 1.) Hypothesis.-Let the equal triangles ABC, DEF, be upon the same side of equal bases BC, EF, in the same straight line BF. Sequence.--The triangles ABC, DEF, shall be between the same parallels. Construction.-1. Join AD; AD shall be parallel to BF. 2. For if it is not, through A draw AG parallel to BF (I. 31), and join GF. |