UNIT III — DISTINCTIO IV: DIVISION OF PLANAR FIGURES AMONG CONSORTS¶
Euclid’s Lost Treatise On Divisions Transmitted & Expanded: Triangles, Centroids, Parallels, Trapezoids, and Circles
Source Text: fibonacci:pg-4, euclid:divisions (Boncompagni Vol. II, pp. 110–148 · fol. 67v–92v) · Latin (1220 / Boncompagni 1862) · Summary · GEOMETOR Notes
Dividing Triangles from a Vertex or Given Boundary Point¶
Latin Locus: De Divisione Trigonorum per Punctum in Latere (pp. 110–112) · Commentary: Notes on Euclid’s On Divisions
Division from a Vertex into Any Given Ratio: To divide a triangle .abg. from vertex .a. into two, three, or more equal or proportional shares, divide the opposite base .bg. in the same ratio [VI. 10] and draw straight lines from .a. to the points of division; for triangles under the same altitude have the same ratio to one another as their bases. [VI. 1]
Equiangular Area-Ratio Lemma: Before dividing from a point that is not a vertex, Leonardo proves from Euclid [VI. 23] that if two triangles share an equal angle at .g., the ratio of their areas is compounded of the ratios of the sides containing that equal angle—that is, $(gd \cdot gz) : (gb \cdot ga)$.
Bisection from an Arbitrary Point on a Side (On Divisions Prop. 4): Suppose two consortes wish to bisect a triangular field ABG by a straight boundary starting from a well, tree, or boundary stone at D on the base BE, where D is not the midpoint.
Construction: Bisect the base at E and join the median AE (which halves ABG). Join AD, and through E draw EZ parallel to AD, [I. 31] meeting side GZ at Z. Draw the dividing boundary DZ.
Proof: Triangles $ADE$ and $ADZ$ stand upon the same base AD and between the same parallels AD and EZ; therefore they are equal in area. [I. 37] Adding the common triangle $EDG$ (or replacing $ADE$ with $ADZ$), the triangular share DZG equals the median half-triangle $AEG$, leaving the quadrilateral share ABDZ equal to the other half!
Numerical Rule: Because EZ $\parallel$ AD, $GD : GE = GA : GZ$. [VI. 2] In the $(13, 14, 15)$ triangle with $BD = 5$, $DG = 9$, $GE = 7$, and $GA = 15$, we have $GZ = \frac{7 \times 15}{9} = 11\frac{2}{3}$ perticae, and both DZG and ABDZ contain exactly $42$ superficial perticae.
Concurrency of Medians & Three-Consort Centroid Division¶
Latin Locus: Preparatoria de Concursu Linearum et Trium Consortum Divisione (pp. 112–120) · Commentary: Notes on Centroid X(2) & Interior Points
Preparatoria: In every triangle ABG, the three straight lines drawn from the angles to the midpoints E, Z, and T of the opposite sides concur at a single point D, where the segment toward each vertex is double the segment toward the midpoint ($AD = 2\,DE$, $BD = 2\,DZ$, $GD = 2\,DT$).
Joining $EZ$ parallel to $AB$, $AB = 2\,EZ$; [VI. 2] hence in the alternate-angle similar triangles $ADB$ and $EDZ$, AD $= 2\,DE$ and BD $= 2\,DZ$. Repeating the same proof for median GT shows it must cut AD at the very same two-thirds point D.
Partition Among Three Consorts: If three consortes wish to share a triangular field ABG with equal access to a central well or landmark at the centroid D, draw the three boundaries AD, BD, and GD. Because $DE = \frac{1}{3} AE$, triangle DBG is one-third of ABG; likewise DAB and DGA are each one-third—so all three shares are equal ($28$ superficial perticae each in the $(13, 14, 15)$ field).
Through Any Given Interior or Exterior Point (pp. 113–117): If the assigned well $P$ is not the centroid (or lies outside the field), Leonardo reduces the straight-line cut through $P$ cutting off a given fraction of area to hyperbolic application of areas [VI. 28–29], solving the resulting quadratic equation by Euclid II.5–6.
Dividing a Triangle Parallel to Its Base¶
Latin Locus: Divisio Trigoni per Lineam Equidistantem Basi (p. 119) · Commentary: Notes on Quadratic Scaling & Trapezoids
Geometric Construction & Proof: Since EZ $\parallel$ BG, triangle AEZ is similar to ABG, so their areas stand in the duplicate ratio of their homologous sides: $\triangle ABG : \triangle AEZ = BA^2 : EA^2$. [VI. 19] To make $\triangle AEZ = \frac{1}{2}\triangle ABG$, we require $BA^2 = 2\,EA^2$, i.e., $EA$ must be the mean proportional between $BA$ and $\frac{1}{2} BA$.
Extend BE beyond A by segment AD $= \frac{1}{2} BA$. Describe the semicircle (O1B) upon diameter $BD$, and erect at A the perpendicular AK to meet the semicircle at K. By the semicircle mean-proportional property [VI. 13], $AK^2 = BA \cdot AD = \frac{1}{2} BA^2$. Transfer length AK by circle (AK) onto side BE at E (so $AE = AK$), and draw EZ parallel to BG meeting GZ at Z. Then AEZ and the residual trapezoid EBGZ are equal!
Surd Example on $(10, 14, 12)$: Let $AB = 10$, $BG = 14$, and $AG = 12$. Then $AD = 5$, and the squares of the cut sides are exact integers even though the lengths themselves are quadratic surds:
$$AE^2 = 10 \times 5 = 50, \qquad AZ^2 = \frac{1}{2} \times 144 = 72, \qquad EZ^2 = \frac{1}{2} \times 196 = 98.$$
Divisions of Quadrilaterals, Trapezoids & Circles¶
Latin Locus: De Divisione Quadrilaterorum et Circulorum (pp. 126–148) · Commentary: Notes on the RMS Trapezoid Invariant
Trapezoid Divided Parallel to Its Bases (Caput Abscisum): Let a trapezoid have upper base (caput) $a$ and lower base (basis) $b$. Completing the flank sides to their apex cone reduces the trapezoid to the difference of two similar triangles proportional to $b^2$ and $a^2$. [VI. 19] Hence a parallel line of length $c$ dividing the trapezoid into upper and lower shares in ratio $p : q$ satisfies the weighted root-mean-square law:
$$c^2 = \frac{q\,a^2 + p\,b^2}{p + q}, \qquad \text{and for bisection } (p = q = 1): \quad c = \sqrt{\frac{a^2 + b^2}{2}}.$$
In the rational gromatic case $a = 7$, $b = 17$, the bisecting parallel chord is $c = \sqrt{\frac{49 + 289}{2}} = \sqrt{169} = 13$, cutting the altitude in ratio $(13 - 7) : (17 - 13) = 6 : 4 = 3 : 2$.
General Quadrilaterals & Polygons: Divided from any vertex or boundary point by first computing the total area via diagonals, isolating the triangle or sub-polygon containing the cut, and applying the side-point construction above.
Circles, Semicircles & Segments (Euclid, On Divisions, Props. 28–29): Circles are divided into equal rings by concentric circles whose radii follow the semicircle mean-proportional law ($r_k^2 = \frac{k}{n} R^2$), or divided by parallel chords separating a given circular segment plus or minus its subtended triangle.