Unit III: Book II — Geometric Algebra & The Golden Section

Coloured Rectangles, Gnomons & Extreme and Mean Ratio (Pickering pp. 51–71, Leaves 85–105)

Book II Definitions: Contained Rectangles & The Gnomon (p. 51)

  • Definition I. Every right-angled parallelogram is said to be contained (×) by any two of the straight lines which contain one of the right angles.

  • Definition II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a gnomon.

[Folio II.4 | p. 56]

Proposition IV. Theorem (Binomial Square, pp. 56–57 · G.147)

If a straight line be divided into any two parts ( + ), the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts.

Book II, Prop. IV — Square of a Divided Line < D K G H > < G F B C > < H G C A > < K E F G > / A C / / C B / / D G / / G B / / H F / / C K / A = (0, 5) C = (3, 5) B = (5, 5) D = (0, 0) E = (5, 0) K = (3, 0) H = (0, 3) G = (3, 3) F = (5, 3)
**Book II, Prop. IV (pp. 56–57).** $(a + b)^2 = a^2 + b^2 + 2ab$ made immediately legible by half-filled diagonal squares and solid yellow complements.

Demonstration (:just:`I.46`, :just:`I.31`, :just:`I.29`, :just:`I.6`, :just:`I.43`).

Along diagonal + , the two parallelograms about the diameter are equilateral right-angled squares ( = 2 and = 2), while the two complements and are equal ([I.43]) and each equals × .

∴ whole square = + + 2 × . Q.E.D.

[Folio II.5 | p. 58]

Proposition V. Theorem (Equal & Unequal Division, p. 58 · G.148)

If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on the half.

Book II, Prop. V — Rectangle of Unequal Segments Plus Inner Square < A C H K > < C D L H > < L M F G > < H L G E > < D B M L > / A C / / C D / / D B / / E B / A = (0, 3) C = (3, 3) D = (5, 3) B = (6, 3) E = (3, 0) F = (6, 0) H = (3, 2) L = (5, 2) M = (6, 2) G = (5, 0) K = (0, 2)
**Book II, Prop. V (p. 58).** $AD \cdot DB + CD^2 = CB^2$ via :term:`gnomon ` equivalence.

Demonstration (:just:`I.43`, :just:`C.N.2`).

Left wing equals vertical complement . Adding to each transforms the rectangle of unequal parts into the L-shaped gnomon; adding inner completes the square on the half-line! Q.E.D.

[Folio II.11 | p. 66]

Proposition XI. Problem (The Golden Section by Colored Areas, p. 66 · G.154)

To divide a given straight line () into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Book II, Prop. XI — Golden Section by Colored Areas < A F G H > < A H K C > < H B D K > / A H / / H B / / E B / / C F / / G K / A = (0, 1) B = (1, 1) C = (0, 0) D = (1, 0) E = (0, 1/2) F = (0, 1/2 + sqrt(5)/2) H = (-1/2 + sqrt(5)/2, 1) G = (-1/2 + sqrt(5)/2, 1/2 + sqrt(5)/2) K = (-1/2 + sqrt(5)/2, 0)
**Book II, Prop. XI (p. 66).** Pure $\mathbb{Q}(\phi)$ area proof of the :term:`Golden Section ` prior to Book V proportion theory.

Construction. Describe the unit square on the given line, bisect the left vertical side at $E$ ([I.10]), join oblique secant ([I.post.1]), step off $EF = EB$ upward ([I.3]), and describe the golden square ([I.46]), partitioning the main square into shared strip and right rectangle .

Demonstration.

By II.6 applied to the bisected vertical side and [I.47] on the right triangle containing :

( + ) = ( + ).

Subtracting the common strip from both ([C.N.3]),

∴ = , i.e. 2 = whole line × . Q.E.F.