Now, since the several bases CB, BG, GH are all equal, the triangles ACB, ABG, AGH are also all equal [I. xxxviii.]. Therefore the triangle ACH is the same multiple of ACB that the base CH is of the base CB. In like manner, the triangle ACL is the same multiple of ACD that the base CL is of the base CD; and it is evident that [I. xxxviii.] if the base HC be greater than CL, the triangle HAC is greater than CAL; if equal, equal; and if less, less. Now we have four magnitudes: the base BC is the first, the base CD the second, the triangle ABC the third, and the triangle ACD the fourth. We have taken equimultiples of the first and third, namely, the base CH, and the triangle ACH; also equimultiples of the second and fourth, namely, the base CL, and the triangle ACL; and we have proved that according as the multiple of the first is greater than, equal to, or less than the multiple of the second, the multiple of the third is greater than, equal to, or less than the multiple of the fourth. Hence [V. Def. v.] the base BC : CD :: the triangle ABC : ACD. 2. The parallelogram EC is double of the triangle ABC [I. xxxiv.], and the parallelogram CF is double of the triangle ACD. Hence [V. xv.] EC : CF :: the triangle ABC : ACD; but ABC : ACD :: BC : CD (Part I.). Therefore [V. xi.] EC : CF :: the base BC : CD. Or thus: Let A, A′ denote the areas of the triangles ABC, ACD, respectively, and P their common altitude; then [II. i., Cor. 1], A = P.BC, A′ = P.CD. Hence = , or A : A′ :: BC : CD. In extending this proof to parallelograms we have only to use P instead of P. PROP. II.—Theorem. If a line (DE) be parallel to a side (BC) of a triangle (ABC), it divides the remaining sides, measured from the opposite angle (A), proportionally; and, conversely, If two sides of a triangle, measured from an angle, be cut proportionally, the line joining the points of section is parallel to the third side. 1. It is required to prove that AD : DB :: AE : EC. Dem.—Join BE, CD. The triangles BDE, CED are on the same base DE, and between the same parallels BC, DE. Hence [I. xxxvii.] they are equal, and therefore [V. vii.] the triangle ADE : BDE :: ADE : CDE; but ADE : BDE :: AD : DB [i.], and ADE : CDE :: AE : EC [i.]. Hence AD : DB :: AE : EC. 2. If AD : DB :: AE : EC, it is required to prove that DE is parallel to BC. Dem.—Let the same construction be made; then AD : DB :: the triangle ADE : BDE [i.]. and AE : EC :: the triangle ADE : CDE [i.]; but AD : DB :: AE : EC (hyp.). Hence ADE : BDE :: ADE : CDE. Therefore [V. ix.] the triangle BDE is equal to CDE, and they are on the same base DE, and on the same side of it; hence they are between the same parallels [I. xxxix.]. Therefore DE is parallel to BC. Observation.—The line DE may cut the sides AB, AC produced through B, C, or through the angle A; but evidently a separate figure for each of these cases is unnecessary. Exercise.