Dem.—Now it can be proved, as in II. vi., that the parallelogram BD is equal to the gnomon OHJ; that is, equal to the difference between the parallelograms PD and PC, or the difference (const.) between KN and PC; that is (const.), equal to X, and BD is escribed to the triangle ABC, and has an angle common with the external angle B. Hence the thing required is done. Observation.—The enunciations of the three foregoing Propositions have been altered, in order to express them in modern technical language. Some writers recommend the student to omit them—we think differently. In the form we have given them they are freed from their usual repulsive appearance. The constructions and demonstrations are Euclid’s, but slightly modified. PROP. XXX.—Theorem. To divide a given line (AB) in “extreme and mean ratio.” Sol.—Divide AB in C, so that the rectangle AB.BC may be equal to the square on AC [II. xi.] Then C is the point required. Dem.—Because the rectangle AB.BC is equal to the square on AC,AB : AC :: AC : BC [xvii.]. Hence AB is cut in extreme and mean ratio in C [Def. ii.]. Exercises. 1. If the three sides of a right-angled triangle be in continued proportion, the hypotenuse is divided in extreme and mean ratio by the perpendicular from the right angle on the hypotenuse. 2. In the same case the greater segment of the hypotenuse is equal to the least side of the triangle. 3. The square on the diameter of the circle described about the triangle formed by the points F, H, D (see fig. II. xi.), is equal to six times the square on the line FD. PROP. XXXI.—Theorem. If any similar rectilineal figure be similarly described on the three sides of a right-angled triangle (ABC), the figure on the hypotenuse is equal to the sum of those described on the two other sides. Dem.—Draw the perpendicular CD [I. xii.]. Then because ABC is a right-angled triangle, and CD is drawn from the right angle perpendicular to the hypotenuse; BD : AD in the duplicate ratio of BA : AC [viii. Cor. 4]. Again, because the figures described on BA, AC are similar, they are in the duplicate ratio of BA : AC [xx.]. Hence [V. xi.] BA : AD :: figure described on BA : figure described on AC. In like manner, AB : BD :: figure described on AB : figure described on BC. Hence [V. xxiv.] AB : sum of AD and BD :: figure described on the line AB : sum of the figures described on the lines AC, BC; but AB is equal to the sum of AD and BD. Therefore [V. a.] the figure described on the line AB is equal to the sum of the similar figures described on the lines AC and BC. Or thus: Let us denote the sides by a, b, c, and the figures by α, β, γ; then because the figures are similar, we have [xx.] α : γ :: a2 : c2 therefore = . In like manner, = ; therefore = ; but a2 + b2 = c2 [I. xlvii.]. Therefore α + β = γ; that is, the sum of the figures on the sides is equal to the figure on the hypotenuse. Exercise.