Book of Lemmas (Liber Assumptorum)¶
Kitāb al-Maʾkhūdhāt / Liber Assumptorum — Archimedes, tr. Thābit ibn Qurra, ed. Borelli & Ecchellensis (ecchellensis-borelli:1661), T. L. Heath (heath:1897, pp. 301–318)
Note
Edition Track (English): Complete 15-proposition planar treatise from heath:1897 (pp. 301–318, archimedes:liber-assumptorum), verified page-by-page across all 18 rendered 300 DPI pages of Heath (1897) and collated with Heiberg's Teubner text (heiberg:1881, Vol. II, pp. 426–445). Switch tracks above for the Summary or Notes.
Propositions 1–3: Tangent Circles, Semicircle Tangents & Chord Ratios¶
Proposition 1 (Homothetic Chords of Tangent Circles): If two circles touch at A, and if CD, EF be parallel diameters in them, then ADF is a straight line (and similarly ACE when tangent externally). Proved by parallel radii and alternate angles (Eucl. I. 29, Eucl. VI. 32).
Proposition 2: Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in T. If now DE be drawn perpendicular to AB, and if AT, DE meet in F, then .
Proposition 3: Let P be any point on a segment of a circle whose base is AB, and let PN be perpendicular to AB. Take D on AB so that . If PQ be an arc equal to the arc PA, and BQ be joined, then BQ, BD are equal (and in the half-angle chord geometry later utilized by Ptolemy).
Propositions 4–6: The Arbelos (Ἄρβηλος) & Twin Circles¶
Proof: By Eucl. III. 31 and Eucl. VI. 8, , so . Since circles and semicircles are to one another as the squares on their diameters (Eucl. XII. 2), the outer semicircle on AB equals the sum of the two inner semicircles on AN, NB together with twice the semicircle (i.e. one whole circle) on PN. Subtracting the two inner semicircles leaves the arbelos equal to the circle on PN!
Proof: Let one circle touch CD at E, the outer semicircle AB at F, and the inner semicircle AC at G, and let its diameter EH be parallel to AB. By Proposition 1, FHG and BFA and AGC align homothetically, yielding ! That is:
Because the expression is symmetric in AC and CB, the second circle inscribed on the opposite side of CD (touching CD, semicircle AB, and semicircle CB) has diameter , strictly equal to EH!
By repeated application of Proposition 1 through the three points of contact, Archimedes proves that for , , and in general if , then:
Furthermore, the height of the centre of this inscribed circle above AB equals its own diameter GH—the base case () of the celebrated ‘ancient proposition’ preserved by Pappus (pappus:synagoge:iv, IV.14: for the -th circle in the infinite tangent chain)!
Propositions 7–13: Neusis Angle Trisection & Orthogonal-Chord Quadrance¶
Proposition 7: If circles be circumscribed about and inscribed in a square, the circumscribed circle is double of the inscribed circle ().
Proof: Draw chord EF parallel to AB, and join OB, OF. Since , the triangle OBC is isosceles (Eucl. I. 5), so . By exterior angles (Eucl. I. 32), . Since and , , and by alternate angles (actually via ): subtends and equals ! Directly: in isosceles triangleOAB, ; then in triangleOAC, exterior . Thus !
Proposition 9: Intersecting chords perpendicular/secant arc-sum lemma.
Proposition 10: Secant and tangent segments from an external point (connected to Apollonius’s Tangencies).
Proposition 11 (Orthogonal Chords Quadrance Invariant):
If two chords AB, CD in a circle intersect at right angles in a point O, not being the centre, then:
\[AO^2 + BO^2 + CO^2 + DO^2 = (\text{diameter})^2 = 4 R^2.\]Proved cleanly via diameter CE perpendicular to AB (or through centre), Eucl. III. 31, and Eucl. I. 47—a pure Rational Trigonometry quadrance identity (geometor:wildberger)!
Proposition 12: Polar-orthocenter property of two tangents to a semicircle (TRperpAB).
Proposition 13: Equal chord segments cut off by perpendiculars onto a diameter ().
Propositions 14 & 15: The Salinon & Pentagon/Decagon Golden Ratio¶
Proof: Because , is the diameter of each lateral semicircle, and , we have . Squaring gives . Converting diameters-squared into semicircles via Eucl. XII. 2 shows that !
Proved from inscribed pentagon/decagon angles ( and of two right angles) and Euclid’s golden-section theorem (Eucl. XIII. 9: the side of the hexagon and the side of the decagon inscribed in the same circle, when added together, cut the whole segment in extreme and mean ratio ).