Sol.—The construction is the same as the last, except that, instead of making the parallelogram KN equal to the excess of the parallelogram PC over the rectilineal figure X, we make it equal to their sum; and then make PI equal to KL; draw IH parallel to AB, and the rest of the construction as before.
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.