Pappus of Alexandria, Mathematical Collection IV §§45-47, pp. 284.21-288.14

 Prop. 45: Division of a angle or circular-arc in a given ratio by a quadratrix Prop. 46: Division of a angle or circular-arc in a given ratio by an Archimedes spiral. Prop. 47: Constructing equal circular-arcs in two unequal circles.

Prop. 45. And so trisecting the given angle or circular-arc is solid, as was previously proved, but cutting the given angle or circular-arc is linear, and this was proved by more recent people, but it will be proved as well by us in two ways.

### Division of angle or circular-arc in a given ratio by Quadratrix

(general diagram) (diagram 1) For let there be a circular-arc LQ of circle KLQ, and let it be required to cut it in the given ratio.

(diagram 2) LBQ is at the center, BK at right angles to BQ, and let there be inscribed squaring line (quadratrix) KADG through K, and (diagram 3) let drawn perpendicular AE be cut at Z, so that as AZ to ZE, so is the given ratio in which want to divide the angle, (diagram 4) and let ZD be parallel to BG, but let BD be joined, (diagram 5) and perpendicular DH. (diagram 6 = general diagram) And so because of the symptom of the line, as AE is to DH, that is to ZE, angle ABG is to angle DBG, therefore by ratio division as AZ is to ZE (AE : DH = angle ABG : DBG => AE-DH : DH = angle ABG - angle DBG : DBG), that is, as the given ratio, so is angle ABD to DBG, i.e., circular-arc LM to MQ.

Note that the basic property of the curve is that AE : DH = arc LMQ : MQ, which is used in this proof. Pappus does not prove this relation as it is obvious from the generation of the curve. Hence, this theorem is an application of the basic symptom of the curve and is not the same as it.

It is obvious that the original motivation for the quadratrix was the division of any angle or circular-arc in this way. It was certainly not the much less evident rectification of the circle, and even less squaring it, despite the name of the curve. However, one can see how this figure would come to suggest a rectification by attending the role of line DH here in the proof that the quadratrix does rectify a circle, for which cf. Pappus, Collectio iv 30-32.

### Division of angle or circular-arc in a given ratio by Archimedes Spiral

(general diagram) (diagram 1) Prop. 46. In another way, circular arc AG of circle AHG is cut. (diagram 2) ABG likewise is at the center, and let spiral BZDG, whose line in the genesis (genetrix) is GB, be inscribed through B, (diagram 3) and let the ratio of DE to EB be the same ratio as the given ratio, (diagram 4) and through E and about center B of let a circular-arc EZ cut the spiral at Z. And let BZ be joined and extended to H (on the outer circle). (diagram 5) Therefore, because of the spiral, as DB is to BZ, that is, BE, so circular-arc AHG is to GH, (diagram 6 = general diagram) and by ratio division as DE is to EB, so circular-arc AH is to HG. But the ratio of DE to EB is the same as the given. Therefore, ratio of circular-arc AH to HG is the same ratio as the given. Therefore, it has been cut.

### Given section of Angle or Circular-arc by any method, cutting off equal arcs

(general diagram)
Prop. 47. From this it is obvious that it is possible from two unequal circles to cut off equal circular-arcs. (analysis) (diagram 1) For let it come to be, and let equals AHB, GQD be taken away, but let the circle about center E be larger. (diagram 2) Therefore, an arc similar to GQD (on circle E) is larger than AHB. (diagram 3) And so let GQ (a smaller arc than GQD) be similar to AHB. (diagram 4) Therefore the ratio of AHB to GQ is given. For the ratio for the whole circular-arc of the circles is the same as the ratio for the diameters of the circles. But GQD is equal to AHB. Therefore, the ratio of GQD to GQ is also given. (diagram 5) And by ratio division. And so it happens that one cuts circular-arc GQD into the given ratio at Q. But this was previously drawn.

Synthesis: (diagram 6) To insure that there are equal arcs, we start with arc GQD in the smaller circle and are required to find an arc on larger circle E equal to GQD. (diagram 7) Let the ratio of the diameter of the larger circle E to the diameter of the smaller circle E be X : Y. (diagram 8) Then with a quadratrix or spiral or whatever, divide arc GQD into arcs GQ and QD such that
arc DQ : arc GQ = (X-Y) : Y
Then by ratio composition
(arc GQ + arc QD) : arc GQ = arc GQD : arc GQ = ((X-Y)+Y) : Y = X : Y
(diagram 9) Hence, arc GQ : arc GQD = diameter of circle E : diameter of the smaller circle.
(diagram 10) Cut off in circle E an arc AHB similar to GQ.

(diagram 11) Since similar arcs are in the ratio of the diameters of the circles,
arc AHB : arc GQ = diameter of circle E : diameter of the smaller circle = arc GQD : arc GQ

Hence, arc AHB = arc GQD.

Note that this proof is a fairly direct conversion of the analysis.