Unit II: Book I — Congruence, Parallels & The Pythagorean Theorem

Showcase Propositions of Book I (Pickering pp. 1–50, Leaves 35–84)

[Folio I.1 | p. 1]

Proposition I. Problem (Equilateral Triangle, p. 1 · G.84)

On a given finite straight line () to describe an equilateral triangle.

**Book I, Prop. I (p. 1).** Exact $\mathbb{Q}(\sqrt{3})$ construction via ``Model.intersect``.

Construction. From the extremities of describe and ([I.post.3]); from their intersection draw and ([I.post.1]).

Demonstration.

= ([I.def.15], radii of ),

= ([I.def.15], radii of ),

∴ = = ([C.N.1]).

∴ is equilateral ([I.def.20]). Q.E.F.

[Folio I.4 | p. 4]

Proposition IV. Theorem (Side-Angle-Side Congruence, p. 4 · G.90)

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, they shall have their bases equal, and the two triangles shall be equal, and their other angles shall be equal, each to each.

**Book I, Prop. IV (p. 4).** Identical color registers encode corresponding sides and subtended angles across both triangles.

Given. Two triangles having equal, equal, and contained apex angles equal.

Demonstration. Superposing the first triangle upon the second so that equal and coincide ([C.N.4]), coincides, whence bases coincide (Axiom X),

∴ are equal and are equal. Q.E.D.

[Folio I.5 | p. 5]

Proposition V. Theorem (Pons Asinorum, p. 5 · G.91)

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall be equal.

**Book I, Prop. V (p. 5).** :term:`Congruence-twin coloring ` turns mirror-symmetric triangles into immediate visual equations.

Construction. Produce equal legs by equal extensions ([I.post.2], [I.3]), and join cross-lines ([I.post.1]).

Demonstration.

+ = + ([C.N.2]), with common apex .

∴ = and lower angles are equal ([I.4]).

In the lower triangles on common base , = ([I.4]).

Subtracting equal parts from equal whole angles ([C.N.3]),

∴ interior base angles = . Q.E.D.

[Folio I.15 & I.32 | p. 15]

Propositions XV & XXXII (Vertical Angles & Angle Sum of a Triangle, pp. 15, 33)

**Book I, Prop. XV (p. 15 · G.104).** Adjacent supplementary pairs cancel common sectors to prove vertical angle equality.

Prop. XV Demonstration (:just:`I.13`, :just:`C.N.3`).

+ = = +

∴ = , and similarly = . Q.E.D.

**Book I, Prop. XXXII (p. 33 · G.125).** Parallel line :byrne-line:`blue solid ` transports both remote interior angles into the exterior angle.

Prop. XXXII Demonstration (:just:`I.31`, :just:`I.29`, :just:`I.13`).

Draw ∥ ([I.31]).

= (alternate, [I.29]), and = (corresponding, [I.29]).

∴ exterior ( + ) = + ,

and adding interior to both yields + + = . Q.E.D.

[Folio I.46 & I.47 | p. 47]

Propositions XLVI & XLVII (The Square & The Pythagorean Theorem, pp. 47–49)

**Book I, Prop. XLVI (p. 47 · G.139).** Construction of the square on :byrne-line:`black solid `.

**Book I, Prop. XLVII (pp. 48–49 · G.140).** Byrne's masterpiece: Euclid's windmill proof rendered as coloured area conservation.

Given. Right-angled triangle with hypotenuse subtending the right angle between legs and .

Construction. Describe and ([I.46]), and divide the hypotenuse square by ([I.31]) into and ; draw windmill chords and ([I.post.1]).

Demonstration.

By collinearity of adjacent right angles ([I.14]), SAS congruence of the rotated windmill triangles ([I.4]), and double-area parallelograms between the same parallels ([I.41]):

= ,

and symmetrically on the right leg (, ):

= .

∴ + = + ([C.N.2]). Q.E.D.