Unit II · Liber Primus Props. VI–XXII: Ceva, Menelaus & Spread Ceva (pp. 7–35)

Original GEOMETOR English translation from Giovanni Ceva’s De lineis rectis se invicem secantibus statica constructio (Milan: Ludovico Monza, 1678, Liber Primus, Props. VI–XXII, pp. 7–35; Tab. 1–3, Figs. 9–27).

1. Proposition VI — Caravaggio Jr.’s Pure Euclidean Proof (Tab. 1, Fig. 10)

Lest strict Euclidean purists object to importing mechanical weights into pure geometry, Giovanni Ceva pauses at Proposition VI (pp. 7–8) to record an elegant synthetic demonstration communicated by his young Milanese colleague Pietro Paolo Caravaggio Jr. (“Petrus Paulus Carauaggius Petri Pauli filius præceptoris sui”, son of Ceva’s Milanese teacher Pietro Paolo Caravaggio Sr.):

Through vertex A of triangle ABC, draw a straight line parallel to base $BC$, meeting the extended cevians $CF$ and $BE$ at $K$ and $L$ (as illustrated in the left panel of Figure 2 in Unit I). By Euclid's similar triangles ([VI.4]):

  1. From $\triangle AKF \sim \triangle BCF$ and $\triangle ALE \sim \triangle CBE$:

    $$\frac{\overline{AF}}{\overline{FB}} = \frac{\overline{AK}}{\overline{BC}}, \qquad \frac{\overline{CE}}{\overline{EA}} = \frac{\overline{BC}}{\overline{AL}}.$$

  2. Since AD passes through the concurrency point P of $BE$ and $CF$, the central ray from $P$ projects $KL$ onto $BC$: $\triangle AKP \sim \triangle DCP$ and $\triangle ALP \sim \triangle DBP$, yielding:

    $$\frac{\overline{BD}}{\overline{DC}} = \frac{\overline{AL}}{\overline{AK}}.$$

Multiplying the three proportions cancels $\overline{AK}$, $\overline{BC}$, and $\overline{AL}$ pairwise, proving $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA} = 1$ by pure Book VI geometry!

2. Propositions VII–VIII & Lemma IV — Compound Ratios, Menelaus & Harmonic Ranges

In Lemma IV and Propositions VII–VIII (pp. 8–12, Tab. 1, Figs. 11–12), Ceva cites Federico Commandino and Christoph Clavius (1589) on the algebra of compound ratios (ratio ex rationibus composita) to state both Ceva’s Theorem and its converse, side-by-side with a two-lever static demonstration of Menelaus's Transversal Theorem (Sphaerica III.1):

  • Proposition VII (Ceva’s Concurrency & Converse): In triangle ABC, three cevians AD, BE, CF concur at a finite (or ideal parallel) point P if and only if the ratio compounded of $(AF : FB)$, $(BD : DC)$, and $(CE : EA)$ is the ratio of equality ($1 : 1$). For if $AD$ and $BE$ meet at $P$, and $CP$ meets $AB$ at $F^*$, then $(AF^* / F^*B) \cdot (BD / DC) \cdot (CE / EA) = 1 = (AF / FB) \cdot (BD / DC) \cdot (CE / EA)$, whence $AF^* / F^*B = AF / FB$ and $F^* = F$ ([V.9]).

  • Proposition VIII (Menelaus’s Transversal & Projective Harmonic Division): Let the straight transversal EF joining two cevian feet E and F be produced until it cuts the extended base lineBC externally at D'. Viewing $\triangle ABC$ cut by transversal EF, static moments equate the unsigned ratio product to $1$, or with signed orientation:

    $$\frac{\overline{AF}}{\overline{FB}} \cdot \frac{\overline{BD'}}{\overline{D'C}} \cdot \frac{\overline{CE}}{\overline{EA}} = -1.$$

A B C D E F P D' Liber Primus Elementum I & Props. VII-VIII: Ceva Concurrency + Menelaus Transversal EF BD/DC = +2/5 and BD'/D'C = -2/5 => Projective Harmonic Cross-Ratio (B, C; D, D') = -1

Dividing Ceva’s equation by Menelaus’s equation immediately yields projective harmonic division (sectio harmonica):

$$\frac{\overline{BD}}{\overline{DC}} = -\frac{\overline{BD'}}{\overline{D'C}} \qquad \Longleftrightarrow \qquad (B, C; D, D') = -1.$$

Thus every concurrent cevian triad $(AD, BE, CF)$ automatically constructs the projective harmonic conjugate D' of D with respect to B and C using a straightedge alone!

3. Propositions IX–XXII — Canonical Triangle Centers & Spread Ceva

Across Propositions IX–XXII (pp. 12–35, Tab. 1–3, Figs. 13–27), Ceva systematically harvests corollaries for parallel transversals, medians, perpendicular altitudes, angle bisectors, and proportional chord configurations. Evaluated over $\mathbb{Q}$ on the canonical Heronian $(13, 14, 15)$ triangle with vertices A $= (5, 12)$, B $= (0, 0)$, and C $= (14, 0)$ ($\text{Area} = 84$, inradius $r = 4$), every classical center (Kimberling ETC) possesses exact integer static weights $(u : v : w)$ and rational coordinates:

  • Incenter X(1) (angle bisectors, $u:v:w = a:b:c = 14:15:13$) at $(6, 4)$;

  • Centroid X(2) (medians, $1:1:1$) at $\left(\frac{19}{3}, 4\right)$;

  • Orthocenter X(4) (altitudes, $70:99:55$) at $\left(5, \frac{15}{4}\right)$;

  • Symmedian Point X(6) ($a^2:b^2:c^2 = 196:225:169$) at $\left(\frac{1673}{295}, \frac{1176}{295}\right)$;

  • Gergonne Contact Point X(7) (Lib. II Prop. I, $\frac{1}{s-a}:\frac{1}{s-b}:\frac{1}{s-c} = 24:28:21$) at $\left(\frac{414}{73}, \frac{288}{73}\right)$;

  • Nagel Point X(8) (isotomic conjugate, $s-a:s-b:s-c = 7:6:8$) at $(7, 4)$—lying with X(1) $= (6, 4)$ and X(2) $= \left(\frac{19}{3}, 4\right)$ along the horizontal Nagel line $y = 4$, while the exterior $A$-Excenter J_A ($-14:15:13$) lies at $(8, -12)$.

A (5,12) B (0,0) C (14,0) D E F X(1) X(2) X(4) X(6) X(7) X(8) Exact Spreads over Q (X(1)): s(BAD) = 16/65, s(DAC) = 16/65 s(CBE) = 4/13, s(EBA) = 4/13 s(ACF) = 1/5, s(FCB) = 1/5 Spread Product = 1 | Quadrance Product = 1 Heronian Triangle (13, 14, 15): Exact Rational Centers X(1), X(2), X(4), X(6), X(7), X(8) Spread Ceva Theorem: [s(BAD)/s(DAC)] * [s(CBE)/s(EBA)] * [s(ACF)/s(FCB)] = 1 (Zero Radicals)

Furthermore, in Rational Trigonometry (Wildberger 2005), squaring the trigonometric form of Ceva’s Theorem replaces transcendental sines with exact rational vertex spreads (sA, sB, sC) and quadrances (Spread Ceva):

$$\frac{s(\angle BAD)}{s(\angle DAC)} \cdot \frac{s(\angle CBE)}{s(\angle EBA)} \cdot \frac{s(\angle ACF)}{s(\angle FCB)} = 1, \qquad \frac{Q(AF)}{Q(FB)} \cdot \frac{Q(BD)}{Q(DC)} \cdot \frac{Q(CE)}{Q(EA)} = 1.$$