PROPOSITION 36. THEOREM. Parallelograms on equal bases, and between the same parallels, are equal in area. Let ABCD, EFGH be parallelograms on equal bases BC, FG, and between the same parallels AH, BG: then shall the parallelogram ABCD be equal to the parallelogram EFGH. therefore BE and CH, which join them towards the same parts, are also equal and parallel. Therefore EBCH is a parallelogram. I. 33. Def. 26. Now the parallelogram ABCD is equal to EBCH; for they are on the same base BC, and between the same parallels BC, AH. I. 35. Also the parallelogram EFGH is equal to EBCH; for they are on the same base EH, and between the same parallels EH, BG. I. 35. Therefore the parallelogram ABCD is equal to the parallelogram EFGH. From the last two Propositions we infer that : Q. E. D. Ax. 1. (i) A parallelogram is equal in area to a rectangle of equal base and equal altitude. (ii) Parallelograms on equal bases and of equal altitudes are equal in area. (iii) of two parallelograms of equal altitudes, that is the greater which has the greater base; and of two parallelograms on equal bases, that is the greater which has the greater altitude. PROPOSITION 37. THEOREM. Triangles on the same base, and between the same parallels, are equal in area. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels BC, AD. Then shall the triangle ABC be equal to the triangle DBC. Construction. Through B draw BE parallel to CA, to meet DA produced in E; I. 31. through C draw CF parallel to BD, to meet AD produced in F. Proof. Then, by construction, each of the figures EBCA, DBCF is a parallelogram. And EBCA is equal to DBCF; Def. 26. for they are on the same base BC, and between the same parallels BC, EF. I. 35. And the triangle ABC is half of the parallelogram EBCA, for the diagonal AB bisects it. I. 34. Also the triangle DBC is half of the parallelogram DBCF, I. 34. for the diagonal DC bisects it. But the halves of equal things are equal; Ax. 7. therefore the triangle ABC is equal to the triangle DBC. [For Exercises see page 73.] Q.E.D. PROPOSITION 38. THEOREM. Triangles on equal bases, and between the same parallels, are equal in area. Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD: then shall the triangle ABC be equal to the triangle DEF. Construction. Through B draw BG parallel to CA, to meet DA produced in G; I. 31. through F draw FH parallel to ED, to meet AD produced in H. Proof. Then, by construction, each of the figures GBCA, DEFH is a parallelogram. And GBCA is equal to DEFH; Def. 26. for they are on equal bases BC, EF, and between the same parallels BF, GH. And the triangle ABC is half of the parallelogram GBCA, for the diagonal AB bisects it. I. 36. I. 34. I. 34. Ax. 7. Also the triangle DEF is half the parallelogram DEFH, for the diagonal DF bisects it. But the halves of equal things are equal: therefore the triangle ABC is equal to the triangle DEF. From this Proposition we infer that : Q.E.D. (i) Triangles on equal bases and of equal altitude are equal in area. (ii) Of two triangles of the same altitude, that is the greater which has the greater base: and of two triangles on the same base, or on equal bases, that is the greater which has the greater altitude. [For Exercises see page 73.] Equal triangles on the same base, and on the same side of it, are between the same parallels. Let the triangles ABC, DBC which stand on the same base BC, and on the same side of it, be equal in area : then shall they be between the same parallels; Join EC. I. 31. Proof. Now the triangle ABC is equal to the triangle EBC, 1 they are on the same base BC, and between the same parallels BC, AE. I. 37. But the triangle ABC is equal to the triangle DBC; Hyp. therefore also the triangle DBC is equal to the triangle EBC; the whole equal to the part; which is impossible. Therefore AE is not parallel to BC. Similarly it can be shewn that no other straight line through A, except AD, is parallel to BC. Therefore AD is parallel to BC. From this Proposition it follows that : Q.E.D. Equal triangles on the same base have equal altitudes. [For Exercises see page 73.] PROPOSITION 40. THEOREM. Equal triangles, on equal bases in the same straight line, and on the same side of it, are between the same parallels. Let the triangles ABC, DEF which stand on equal bases BC, EF, in the same straight line BF, and on the same side of it, be equal in area : then shall they be between the same parallels; that is, if AD be joined, AD shall be parallel to BF. Construction. For if AD be not parallel to BF, if possible, through A draw AG parallel to BF, Join GF. I. 31. Proof. Now the triangle ABC is equal to the triangle GEF, for they are on equal bases BC, EF, and between the same parallels BF, AG. I. 38. But the triangle ABC is equal to the triangle DEF: Hyp. therefore also the triangle DEF is equal to the triangle GEF : the whole equal to the part; which is impossible. Therefore AG is not parallel to BF. Similarly it can be shewn that no other straight line through A, except AD, is parallel to BF. Therefore AD is parallel to BF. From this Proposition it follows that: Q.E.D. (i) Equal triangles on equal bases have equal altitudes. (ii) Equal triangles of equal altitudes have equal bases. |
