halves of equal things are equal. [Ax. 7.] Therefore the triangle ABC is equal to the triangle DBC. Therefore, triangles, &c. Q.E.D. Ex. If AC and DB cross each other at O in such a way that AB is parallel to CB; shew that the triangle AOD is equal to the triangle BOC. PROPOSITION 38. THEOREM. Triangles on equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be on equal bases 1. BC, EF and between the same parallels BF, AD: the triangle ABC shall be equal to the triangle DEF. 2. B N E Produce AD both ways to the points G, H; through B draw BG parallel to CA, and through F draw FH parallel to ED. [I. 31.] 3. 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]. And the triangle ABC is half of the parallelogram GBCA, because the diameter AB bisects it; [I. 34.] and the triangle DEF is half of the parallelogram DEFH, because the diameter DF bisects it. But the halves of equal things are equal. [Ax. 7.] Therefore the triangle ABC is equal to the triangle DEF. Therefore, triangles, &c. Q.E.D. An important case of prop. 38 is when the two triangles have their vertices terminated in the same point. Ex. The four triangles into which a parallelogram is divided by its diagonals are equal to one another- [See Ex. 3 to prop. 34.] PROPOSITION 39. THEOREM. Equal triangles on the same base, and on the same sides of it, are between the same parallels. Let the equal triangles ABC, DBC be on the 1. same base BC, and on the same side of it: they shall be between the same parallels. 2. Join AD. AD shall be A parallel to BC. For if it is not, through A draw AE parallel to BC, [I. 31.] and join EC. Then the triangle ABC B 3. is equal to the triangle EBC, because 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, [Ax. 1.] the greater to the less; which is impossible. Therefore AE is not parallel to BC. In the same manner it can be demonstrated, that no other line through A but AD is parallel to BC; therefore AD is parallel to BC. Therefore, equal triangles, &c. Q.E.D. Ex. The straight line which joins the middle points of two sides of a triangle is parollel to the base. PROPOSITION 40. THEOREM. Equal triangles, on equal bases, in the same straight Line, and towards the same parts, are between the same parallels. 1. Let the equal triangles ABC, DEF be on equal bases BC, EF, in the same straight line BF, and towards the same parts: they shall be between the same parallels. AD Join AD. 2. shall be parallel to BF. For if it is not, through A draw AG parallel to BF [I. 31.] and join GF. B Then the triangle ABC is equal to the triangle 3. GEF; because they are on equal bases BC, EF, and between the same parallels. [I. 38.] But the triangle ABC is equal to the triangle DEF. [Hyp.] Therefore also the triangle DEF is equal to the triangle GEF, [Ax. 1.] the greater to the less; which is impossible. Therefore AG is not parallel to BF. In the same manner it can be demonstrated that no other line through A but AD is parallel to BF; therefore AD is parallel to BF. Therefore, equal triangles, &c. Q.E.D. 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 1. EBC be on the same base BC, and between the same parallels BC, AE: the parallelogram ABCD shall be double of the triangle EBC. parallelogram. [I. 34.] Therefore the parallelogram ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c. Q.E.D. Hence the area of a triangle is half that of a rectangle having the same base and altitude. PROPOSITION 42. PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. A. Let ABC be the given triangle, and D the given rectilineal angle: it is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Bisect BC in E: [I. 10.] Join AE, and at 2. the point E in the straight line EC make the angle CEF equal to D; [I. 23.] through A draw AFG parallel to EC, and through C draw CG parallel to EF. [I. 31.] Therefore FECG is a parallelogram. [Def.] And because BE is equal to EC, [Const.] the 3. triangle ABE is equal to the triangle AEC, because they are on equal bases BE, EC, and between the same parallels BC, AG. [I. 38.] Therefore the triangle ABC is double of the triangle AEC. But the parallelogram FECG is also double of the triangle AEC, because they are on the same base EC, and between the same parallels EC. AG. [I. 41.] Therefore the parallelogram FECG is equal to the triangle ABC; [Ax. 6.] and it has one of its angles CEF equal to the given angle D. [Const.] Therefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F. Ex. Construct a square which shall be equal to a given rightangled isosceles triangle. PROPOSITION 43. THEOREM. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF parallelograms about AC, that is, through which AC passes; and BK, KD the other parallelograms which make up the whole figure ABCD, and which are therefore called the complements: the complement BK shall be equal to the complement KD. Because ABCD A 3. is a parallelogram, E B C H K F and AC its diameter, the triangle ABC is equal to the triangle ADC. [I. 34.] Again, because AEKH is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle AHK. [I. 34.] For the like reason the triangle KGC is equal to the triangle KFC. Therefore, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC. [Ax. 2.] But the whole triangle ABC was proved equal to the whole triangle ADC. Therefore the remainder, the complement BK, is equal to the remainder, the complement KD. [Ax. 3.] Therefore, the complements, &c. Q.E.D. Ex. The parallelogram EH is equiangular to the parallelogram BD. |