Pappus of Alexandria, Mathematical Collection V §§11-13, pp. 234.22-242.12:

Prop. 11: Amazing proof that: The circular-arcs of circles are to one another as their diameters

For those who do not like the amazing proof:

Prop. 12: A circle has the same ratio to a section as the circumference of the circle to the circular arc of the section.

Prop. 13: Part I: Similar segments of circles are to one another as the squares of their bases are to one another. Part II: as their circular-arcs are to one another, so are their bases.

Prop. 14: If two radii in one triangle form equal angles with radii in another, then the triangles formed by a tangent from one radius meeting the extension of the other and the perpendicular from the tangent to the other radius (half-chords) will be as the squares of the half chords.

Prop. 11:

This too is of the same examination as those we just mentioned, the semicircle is the largest of circular segments having equal circular-arcs. We will prove this after first inscribing (proving) some things assumed for it.

The circular-arcs of circles are to one another as their diameters. (diagram 1) Let there be two circles AB, GD, and their diameters AB, GD. I say that as the circular-arc of circle AB is to the circular-arc of circle GD, so is straight-line AB to GD.

(diagram 2) Since as circle AB is to circle GD, so is the square of AB to the square of GD, (diagram 3) but the rectangle enclosed by straight-line AB and the circular-arc of circle is four-times circle AB, but the rectangle enclosed by straight-line GD and the circular-arc of circle GD is four-times circle GD (for this have been proved). And, therefore, as the rectangle of AB and the circular-arc of circle AB to the rectangle of GD and the circular-arc of circle GD, so is the square of AB to the square of GD. And by proportion alteration as the rectangle of the circular-arc of circle AB and AB is to the square of AB, so is the rectangle of the circular-arc of circle GD and GD is to the square of GD. Therefore as the circular-arc of circle AB is to AB, s is the circular-arc of GD to GD. And by proportion alteration as the circular-arc of AB is to the circular-arc of GD, so is AB to GD.

Prop. 12: This is proved even without assuming that the rectangle of the diameter of the circle and its circular-arc is four-times the cirlce. For, similar polygons inscribed in the circle, or circumscribed, have perimeters in the same ratio to one another as the lines from the center (radii), so that even the circular arcs of the circles are to one another as the diameters.

(diagram 1)Again let there be a circle ABG about center D, and its radius is DB, and from D let some line be drawn, DE. I say that as perimeter ABG of the circle to circular-arc BZE, so is circle ABG to section BDE. (diagram 2)And so if circular-arc BZE is commensurable with perimeter ABG, then let perimeter ABG of the circle be divided into measures and from the points of the division let straight-lines be joined to center D. Since all the segments will be congruent with one another, and the number of them is equal to the number of the measures, therefore, as the whole perimeter ABG is to the section BDE, so is circle ABG to section BDE [15 of Elements 5].

(diagram 3)If it is not commensurable with the circular-arc, it is likewise the case that as circle ABG is to section BDE, so is perimeter ABG to circular-arc BZE. (diagram 4)Let it be the case, if possible, that as circle ABG is to section BDE, so is its perimeter ABG to circular-arc BZ which is first less than circular-arc BZE. And let some other circular-arc BH be taken which is larger than BZ but less than BZE, while being commensurable with perimeter ABG, as is a lemma of the Sphaerica (found in a scholion to Theodosius Sphaerica iii 9). And let DH be joined. And so, because of the previous claims, it is also that as circle ABG is to section BDH, so is perimeter ABG of the circle to the circular-arc BZH. But perimeter ABG of the circle to circular-arc BZH has a smaller ratio than it does to circular-arc BZ, that is than circle ABG has to section BDE. And so, circle ABG will have a smaller ratio to BDH than it has to section BDE, which is absurd. Therefore, it is not the case that circle ABG is to section BDE, so is its perimeter ABG to circular-arc BZ, which is less than circular-arc BZE.

(diagram 5)I say, in fact, that it is not to a larger than BZE either. For if it is possible, let it be to circular-arc BEG, (diagram 6)and let some circular-arc BEQ likewise be taken which is larger than circular-arc BZE but smaller than circular-arc BEG, while being commensurate with perimeter ABG of the circle, and let DQ be joined. And so since again as circle ABG is to section BDQ, so is perimeter ABG of the circle to circular-arc BEQ. But perimeter ABG has a larger ratio to circular-arc BEQ than it has to circular-arc BEG, that is than circle ABG to section BDE, it clearly follows that circle ABG will have a larger ratio to section BDQ than it has to section BDE, which is absurd. Therefore, it is not the case that as circle ABG to section BDE, so is its perimeter ABG to a circular-arc larger than BZE.

But it was proved that it is not to a smaller either. Therefore, as circle ABG is to section BDE, so is its perimeter ABG to circular-arc BZE.

Prop. 13. Similar segments of circles are to one another as the squares of their bases are to one another and as their circular-arcs are to one another, so are their bases. (diagram 1) For let there be similar segments of circles, ABG, DEZ. I say that as segment ABG is to DEZ, so is the square of AG to the square of DZ, and as circular-arc ABG is to DEZ, so is AG to DZ.

(diagram 2) Let the circles be completed and let their centers be taken, H, Q, and let AHG, DQZ be joined. (diagram 3) And so since segments ABG, DEZ are similar, the angle at H is equal to the angle at Q, and triangle AHG is similar to DQZ, and circular-arc ABG is similar to DEZ. (diagram 4) Therefore, as circle ABG is to section AHGB, so is the perimeter of circle ABG to circular-arc ABG, that is 4 right-angles to angle H. But as circle DEZ is to section DQZE, so is the perimeter of circle DEZ is to circular-arc DEZ, that is 4 right-angles to angle Q. And angle Q is equal to angle H. Therefore, as circle ABG is to section AHGB, so is circle DEZ to section DQZE, and by alternating proportion as circle ABG is to DEZ, so is section AHGB to section DQZE. (diagram 5) But as the circle is to the circle, so is the square of AH to the square of DQ, (diagram 6) that is triangle AHG to triangle DQZ. Therefore, as section AHGB is to section DQZE, so is triangle AHG to triangle DQZ. (diagram 7) And by proportion subtraction (Euclid Elements v 19), segment ABG is to segement DEZ as triangle AHG is to triangle DQZ, (diagram 8) that is as the square of AG is to the square of DZ.

(diagram 7) I say that it is also the case that as circular-arc ABG is to DEZ, so is AB to DZ.

For when the same things are constructed, as the circular-arc of circle ABG is to the circular-arc of circle DEZ, so is circular-arc ABG to DEZ. But the circular-arcs of the circles are to one another as AH is to DQ, that is AG to DZ. Therefore, as circular-arc ABG is to DEZ, so is AG to DZ.

Prop. 14. (diagram 1)Let there be two circles, and at their centers let there be equal angles enclosed by ABG, DEZ, and tangents AH, DQ, perpendiculars AK, DL. To show that as triangle AHK is to triangle AGK, so is triangle DQL to triangle DZL.

(diagram 2) It is obvious from what was previously inscribed (proved). For triangle AHK becomes similar to DQL, and trilateral AGK to trilateral DZL, and each has a respective ratio to one another which the square of AK has to the square of DL.