UNIT II — DISTINCTIO III: MENSURATION OF FIELDS, HERON’S PROOF & CIRCLE BOUNDS

Scalene Altitudes, the Euclidean Incenter Proof of Heron’s Formula, Quadrilaterals, and the 96-gon Circle Measurement

Source Text: fibonacci:pg-3 (Boncompagni Vol. II, pp. 30–110 · fol. 18v–67v) · Latin (1220 / Boncompagni 1862) · Summary · GEOMETOR Notes

[Folio fol. 18v–24r | p. 30–39]

Triangles by Cathetus and Casus

Latin Locus: De Mensuratione Trigonorum (pp. 30–39) · Commentary: Notes on (13, 14, 15) & Rational Trigonometry

Here begins the Third Distinction, on the measurement of all planar fields (in mensuratione omnium camporum); and first concerning triangles, which are the primary elements into which every rectilinear figure is resolved.

Every triangle is either orthogonal (right-angled), oxygonal (acute-angled), or amblygonal (obtuse-angled). In every triangle, the surface (embadum) is equal to the product of half its perpendicular altitude (cathetus) into the base on which it falls, or half the base into the whole cathetus. [I. 41]

When the three sides of an acute scalene triangle .abg. are known—for example, $ab = 13$ perticae, $ag = 15$ perticae, and base $bg = 14$ perticae—the surveyor finds first the projection (casus) of the shorter side upon the base by Euclid’s theorem on acute triangles [II. 13]:

[II. 13]Subtract the square of $ab$ ($169$) from the sum of the squares of $bg$ ($196$) and $ag$ ($225$), leaving $252$; divide this remainder by twice the base $bg$ ($28$), yielding the greater casus $dg = 9$. Subtracting $9$ from $14$ leaves the lesser casus $bd = 5$. Then by the theorem of the right triangle [I. 47], the square of the cathetus $ad$ is $ab^2 - bd^2 = 169 - 25 = 144$, whence $ad = 12$ perticae, and the embadum is $\frac{1}{2} \times 14 \times 12 = 84$ superficial perticae.
[Folio fol. 24v–25v | p. 40–41]

Heron’s Semiperimeter Rule & Fibonacci’s Incenter Proof

Latin Locus: Regula et Demonstratio Campi Trigonij (pp. 40–41) · Commentary: Notes on Heron & the Banū Mūsā Transmission

Without seeking the cathetus, to find the area (embadum) of any triangle directly from its three known sides: add the three sides of the triangle together and take half of their sum; from this semiperimeter subtract each side of the triangle in order, keeping the three remainders; multiply the first remainder into the second, that product into the third, and the result into the semiperimeter itself; the square root of this final product is the area of the triangle.

For the triangle of sides $13, 14, 15$, the sum is $42$ and its half is $21$. Subtracting $13, 14, 15$ from $21$ leaves $8, 7, 6$; multiplying $21 \times 8 \times 7 \times 6 = 7{,}056$, its square root is $84$ superficial perticae. Now let this admirable rule of Heron (heron:metrica, banu-musa:verba) be proved geometrically.

Distinctio III — Fibonacci's Geometric Incenter Proof of Heron's Formula ( T H ): (x - 6)**2 + (y - 4)**2 - 16 = 0 [ B L ]: y = 0 [ B E ]: 12*x - 5*y = 0 [ G Z ]: -12*x - 9*y + 168 = 0 [ B T ]: -12*x + 18*y = 0 [ G T ]: -56*x/5 - 112*y/5 + 784/5 = 0 / A T /: length = sqrt(65) / B T /: length = 2*sqrt(13) / G T /: length = 4*sqrt(5) / T H /: length = 4 / T E /: length = 4 / T Z /: length = 4 / G L /: length = 7 / B H /: length = 6 / H G /: length = 8 B = (0, 0) G = (14, 0) A = (5, 12) H = (6, 0) E = (30/13, 72/13) Z = (46/5, 32/5) T = (6, 4) L = (21, 0) B G A H E Z T L
Triangle < A B G > (13, 14, 15) with angle bisectors meeting at incenter T, equal perpendiculars / T H /, / T E /, / T Z / of inradius 4 to the sides, and base [ B L ] extended to L so BL equals the semiperimeter 21.

Let there be given any triangle ABG, and let two of its angles—namely at B and G—be bisected by the straight lines BT and GT meeting at T. [I. 9]

  1. Equal Inradii & Tangency Segments: From T drop perpendiculars TH upon the base, TE upon side BE, and TZ upon side GZ. [I. 12] Because right-angled triangles sharing hypotenuse BT or GT have equal acute angles at the bisected vertices, [I. 26] the three perpendiculars TH, TE, and TZ are all equal to one another, and the circle (TH) touches all three sides. [IV. 4] Joining AT, it likewise bisects the apex angle at A, giving three pairs of equal tangent segments ($BE = BH$, $HG = GZ$, $AE = AZ$).

  2. Extending the Base to the Semiperimeter: Produce the base line BL beyond G to L so that the added segment GL equals $AE$ ($= AZ$). Since $BH + HG + GL = BH + GZ + AE$ is half the sum of all six tangent segments, the extended line $BL$ equals the semiperimeter $s$ of triangle ABG ($BL = 21$). Moreover, the three excesses $(s - a, s - b, s - c)$ appear directly as the segments on $BL$:

    $$GL = s - BG, \qquad BH = s - AG, \qquad HG = s - AB.$$

  3. Area by Triangulation Around the Incenter: Triangle ABG is partitioned into three sub-triangles along AT, BT, and GT, each having altitude equal to TH. Therefore the embadum equals $\frac{1}{2} \cdot TH \cdot (\text{perimeter}) = TH \cdot BL$, and its square is $TH^2 \cdot BL^2$.

  4. Similarity via Cyclic Auxiliary Construction: Erecting perpendiculars at A and B to obtain the supplementary cyclic quadrilateral with angle bisectors meeting at right angles, similar right triangles establish the proportion:

    $$BL : BH = HG : \text{auxiliary segment}, \qquad \text{whence} \quad TH^2 \cdot BL = GL \cdot BH \cdot HG.$$

    Multiplying both sides by $BL$ yields the sought equality:

    $$\Delta^2 = TH^2 \cdot BL^2 = BL \cdot GL \cdot BH \cdot HG = s(s - a)(s - b)(s - c).$$

Which was to be demonstrated (quod oportebat ostendere).
[Folio fol. 34r–53v | p. 56–87]

Quadrilaterals & Multi-Angular Polygons

Latin Locus: De Campis Quadrilateris et Multangulis (pp. 56–87) · Commentary: Notes on Quadrilaterals

Moving from three-sided to four-sided and multi-sided fields, Leonardo classifies every quadrilateral systematically:

  1. Rhombi (Campi parte altera longiores sed non rectanguli / caput abscisi): Measured through half the product of their perpendicular diagonals $e_1, e_2$, since $e_1^2 + e_2^2 = 4a^2$.

  2. Trapezoids (Caput abscisum): Having parallel bases (caput $a$ and basis $b$) with known flank sides $c, d$. Drawing a line parallel to one flank reduces the figure to a parallelogram plus a known scalene triangle, giving the cathetus $h$ and exact area $\frac{a + b}{2} \cdot h$.

  3. General Quadrilaterals & Polygons: Every general quadrilateral or multi-angular field is resolved into triangles by measuring one or more diagonals (diametri), warning surveyors explicitly against the false agrimensor rule of multiplying averages of opposite sides ($\frac{a+c}{2}\cdot\frac{b+d}{2}$), which holds only for rectangles!

[Folio fol. 54r–56v | p. 88–91]

Archimedes’ Circle Measurement Proved with Small Numbers

Latin Locus: De Mensuratione Circulorum (pp. 88–91) · Commentary: Notes on the 96-gon Bounds & 864/275

It must also be shown how it was discovered by the philosopher Archimedes (archimedes:circle) that the circumference of every circle is three times its diameter and a seventh part more—since it is possible to demonstrate most fully with small numbers (cum parvis numeris) those things which he himself showed with large numbers.
Distinctio III — Circle Measurement (1440 : 458 1/3 = 864 : 275) ( C A ): x**2 + y**2 - 225 = 0 [ C G ]: y = 0 [ A F ]: x - 15 = 0 [ C Y ]: -5*sqrt(3)*x + 15*y = 0 [ C F ]: -15*x/2 + y*(15*sqrt(3)/2 + 15) = 0 / E Z /: length = 10*sqrt(3) / C Z /: length = 10*sqrt(3) / Y M /: length = 15 / Y A /: length = 15*sqrt(2 - sqrt(3)) / A M /: length = 15*sqrt(2 - sqrt(3)) / A F /: length = 30 - 15*sqrt(3) / G Y /: length = 15*sqrt(sqrt(3) + 2) B = (0, 15) D = (0, -15) C = (0, 0) A = (15, 0) G = (-15, 0) E = (15, 5*sqrt(3)) Z = (15, -5*sqrt(3)) Y = (15*sqrt(3)/2, 15/2) M = (15*sqrt(3)/2, -15/2) F = (15, 30 - 15*sqrt(3)) B D C A G E Z Y M F
Circle ( C A ) on diameter [ C G ] through center C, with tangent / E Z / at A, 30-degree half-hexagon ray [ C Y ] cutting circle at Y, angle bisector [ C F ] cutting tangent at F, and inscribed chord / Y A / in semicircle triangle GYA.

  1. Circumscribed 96-gon Upper Bound: Let circle (CA) have diameter $GA$ through center C and G, with perpendicular diameter through B and D, and tangent EZ touching the circle at A. [III. 16] Let angle $ECA$ be one-third of a right angle ($30^\circ$), so EZ is the side of the circumscribed regular hexagon and $CE = 2\,EA$. Setting $EA = 15$ (or scaling to clear fractions at $CA = 720$), then $CA = \sqrt{CE^2 - EA^2} = \sqrt{675} < 26$.

    Bisecting angle $ECA$ successively four times by rays such as CF—passing from the $6$-gon to the $12$-gon (AF), $24$-gon, $48$-gon, and $96$-gon—each angle bisector divides the tangent base in the ratio of hypotenuse to radius [VI. 3]:

    $$CA : AF = (CE + CA) : EA.$$

    Carrying out the four bisections with integer-and-fraction bounds scaled to diameter $d = 1440$ ($CA = 720$), Leonardo proves that the perimeter of the circumscribed $96$-gon bears to the diameter a ratio smaller than $1440 : 458\frac{1}{5}$:

    $$\frac{\text{Circumference}}{\text{Diameter}} < \frac{1440}{458\frac{1}{5}} = \frac{7200}{2291} \approx 3.1427324\dots < 3\frac{1}{7}.$$

  2. Inscribed 96-gon Lower Bound: Conversely, in the semicircle on diameter $GA$, let YA be the side of the inscribed hexagon ($YA = \text{radius}$), forming right triangle $GYA$ with leg GY and altitude projections at M (YM, AM). [III. 31] Four successive chord bisections ($6 \to 12 \to 24 \to 48 \to 96$) establish that the inscribed $96$-gon perimeter exceeds the ratio $1440 : 458\frac{4}{9}$:

    $$\frac{\text{Circumference}}{\text{Diameter}} > \frac{1440}{458\frac{4}{9}} = \frac{6480}{2063} \approx 3.1410567\dots > 3\frac{10}{71}.$$

  3. Fibonacci’s Own Mean Ratio ($864 : 275$) & Reduction to $22 : 7$: Choosing the clean harmonic divisor $458\frac{1}{3}$ strictly between $458\frac{1}{5}$ and $458\frac{4}{9}$, Leonardo obtains his refined rational estimate:

    $$\frac{1440}{458\frac{1}{3}} = \frac{4320}{1375} = \frac{864}{275} = 3\frac{39}{275} \approx 3.141818\dots$$

    And diminishing the denominator $275$ by the minute fraction $\frac{1}{11}$ recovers the surveyor’s Archimedean rule:

    $$\frac{864}{275 - \frac{1}{11}} = \frac{9504}{3024} = \frac{22}{7} = 3\frac{1}{7}.$$

    Hence the area (embadum) of any circle equals half its circumference multiplied into half its diameter, or $\frac{11}{14}$ of the square on its diameter; and circular sectors (sectores) and segments (portiones) are measured by subtracting or adding the subtended isosceles triangle.