Archimedis Opera cum Apollonii Conicorum Libri IIII (1675) - Page 23

Prop. I.

It is incorrectly numbered among the common statements.

If a polygon (ABCDEF) is inscribed in a circle (ADF), it is clear that the perimeter of the inscribed polygon is smaller than the circumference of the circle.

Fig. 5. 2 ax. of this.

For each side, as AB, is less than the arc (AB) which it subtends; and consequently, all the sides taken together are less than all the arcs taken together, that is, the whole perimeter of the polygon is smaller than the whole circumference of the circle.

Corollary 1. By the same reasoning, however any arc (AD) is divided and the subtending chords (AB, BC, CD) are drawn, the whole arc is greater than all the subtended chords.

Corollary 2. The sine original: "Sinus rectus" is smaller than its arc, that is, when ZYX is drawn from the center Z perpendicular to AB, AY is less than the arc AX.
For AYB (2 AY) is less than AXB (2 AX).

Prop. II.

Fig. 6.

If a polygon (MNOPQ) is described about a circle (ABCDE), the perimeter of the circumscribed polygon will be greater than the perimeter of the circle.

2 ax. of this.

For the combined line AM + BM is greater than the arc AB, and BN + CN is greater than the arc BC; and so for the rest: wherefore the whole perimeter of the circumscribed figure is greater than the whole circumference of the circle.

Corollary 1. By similar reasoning, however any arc is divided, the circumscribed tangents are greater than the whole arc.

Corollary 2. The tangent is greater than its arc, namely, when ZA, ZM are drawn, AM is greater than AY. For AM + BM (2 AM) is greater than AYB (2 AY).

Prop. III.

Fig. 7.

Given two unequal magnitudes (A, B), it is possible to find two unequal straight lines, such that the greater line has a smaller ratio to the smaller line than the greater magnitude (A) to the smaller (B).

5 ax. of this. constr. 3. 5.

Let A - B be multiplied by some number (say N) until the produced magnitude, which I call X, exceeds B. Then, having taken any straight line R, let R : S

1 : N

A - B : X. I say that R + S and S are the lines sought. For because B is less than X, then A - B : B is greater than (A - B : X, which is) R : S. Whence by compounding, A : B will be greater than R + S : S. Q. E. F.