PROPOSITION 38. THEOREM. Triangles on equal bases, and between the same parallels, are equal in area. F B СЕ 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 since they are on equal bases BC, EF, and between the same parallels BF, GH; .. the parallelogram GBCA = the parallelogram DEFH. I. 36. Now the diagonal DF bisects GBCA; I. 34. .. the triangle DEF is half the parallelogram DEFH. And the halves of equal things are equal. Ax. 7. Therefore the triangle ABC is equal to the triangle DEF. Q.E.D. Def. 36. I. 34. From this Proposition we infer that: (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. PROPOSITION 39. THEOREM. in area Equal triangles on the same base, and on the same side of it, are between the same parallels. A B 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 the triangles ABC, DBC be between the same parallels; that is, if AD be joined, AD shall be parallel to BC. Construction. For if AD be not parallel to BC, if possible, through A draw AE parallel to BC, I. 31. meeting BD, or BD produced, in E. Join EC. Proof. Now the triangles ABC, EBC are on the same base BC, and between the same parallels BC, AE; .:. the triangle ABC = the triangle EBC. I. 37. But the triangle ABC = the triangle DBC; Нур. ... the triangle DBC=the triangle EBC; that is, the whole is equal to a part; which is impossible. .. 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. Q.E.D. From this Proposition it follows that: (For Exercises see page 79.] 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 the triangles ABC, DEF 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, I. 31. meeting ED, or ED produced, in G. Join GF. Proof. Now the triangles ABC, GEF are on equal bases BC, EF, and between the same parallels BF, AG; .. the triangle ABC = the triangle GEF. I. 38. But the triangle ABC=the triangle DEF: Hyp. ... the triangle DEF = the triangle GEF: that is, the whole is equal to a part; which is impossible. .. 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. Q.E.D. = From this Proposition it follows that: (i) Equal triangles on equal bases have equal altitudes. (ii) Equal triangles of equal altitudes have equal bases. EXERCISES ON PROPOSITIONS 37-40. DEFINITION. Each of the three straight lines which join the angular points of a triangle to the middle points of the opposite sides is called a Median of the triangle. a ON PROP. 37. 1. If, in the figure of Prop. 37, AC and BD intersect in K, shew that (i) the triangles AKB, DKC are equal in area. (ii) the quadrilaterals EBKA, FCKD are equal. 2. In the figure of 1. 16, shew that the triangles ABC, FBC are equal in area. 3. On the base of a given triangle construct a second triangle, equal in area to the first, and having its vertex in a given straight line. 4. Describe an isosceles triangle equal in area to a given triangle and standing on the same base. ON PROP. 38. 5. A triangle is divided by each of its medians into two parts of equal area. 6. A parallelogram is divided by its diagonals into four triangles of equal area. 7. ABC is a triangle, and its base BC is bisected at X; if Y be any point in the median AX, shew that the triangles ABY, ACY are equal in area. 8. In AC, a diagonal of the parallelogram ABCD, any point X is taken, and XB, XD are drawn : shew that the triangle BAX is equal to the triangle DAX. 9. If two triangles have two sides of one respectively equal to two sides of the other, and the angles contained by those sides supplementary, the triangles are equal in area. ON PROP. 39. 10. The straight line which joins the middle points of two sides of a triangle is parallel to the third side. 11. If two straight lines AB, CD intersect in O, so that the triangle AOC is equal to the iriangle DOB, shew that AD and CB are parallel. ON PROP. 40. 12. Deduce Prop. 40 from Prop. 39 by joining AE, AF in the figure of page 78. PROPOSITION 41. THEOREM. If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Then shall the parallelogram ABCD be double of the triangle EBC. Join AC. Proof. Now the triangles ABC, EBC are on the same base BC, and between the same parallels BC, AE; .. the triangle ABC=the triangle EBC. I. 37. And since the diagonal AC bisects ABCD; .. the parallelogram ABCD is double of the triangle ABC. Therefore the parallelogram ABCD is also double of the triangle EBC. Q.E.D. I. 34. EXERCISES. 1. ABCD is a parallelogram, and X, Y are the middle points of the sides AD, BC; if Z is any point in XY, or XY produced, shew that the triangle AZB is one quarter of the parallelogram ABCD. 2. Describe a right-angled isosceles triangle equal to a given square. 3. If ABCD is a parallelogram, and X, Y any points in DC and AD respectively : shew that the triangles AXB, BỘC are equal in area. 4. ABCD is a parallelogram, and P is any point within it ; shew that the sum of the triangles PÁB, PCD is equal to half the paral. lelogram. |