v. The altitude of any figure is the length of the perpendicular from its highest point to its base.
vi. Two corresponding angles of two figures have the sides about them reciprocally proportional, when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first.
This is evidently equivalent to saying that a side of the first is to a side of the second in the reciprocal ratio of the remaining side of the first to the remaining side of the second.
PROP. I.—Theorem.
Triangles (ABC, ACD) and parallelograms (EC, CF) which have the same altitude are to one another as their bases (BC, CD).
Dem.—Produce BD both ways, and cut off any number of parts BG, GH, &c., each equal to CB, and any number DK, KL, each equal to CD. Join AG, AH, AK, AL.
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. |