Unit 3: Book IV & VI Extensions, Part A — Ceva, Menelaus, Similitude, Harmonic & Anharmonic Ratios¶
John Casey — A Sequel to the First Six Books of the Elements of Euclid (5th ed., 1888 / 6th ed., 1892)
Book VI, Section I — Theory of Concurrent Lines & Transversals¶
Book VI, Section I, Proposition 2 (:person:`Ceva’s <ceva>` Theorem, 1678). If three straight lines drawn from the vertices $A, B, C$ of a triangle to meet the opposite sides in $D, E, F$ be concurrent at a point $O$, the product of three alternate segments of the sides is equal to the product of the other three:
$$ \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = +1; $$
and conversely, if this signed ratio relation holds, the three cevians $AD, BE, CF$ are concurrent (or parallel).
Demonstration. By Euclid VI.1, triangles of equal altitude are to one another as their bases; hence $BD : DC = \triangle AOB : \triangle AOC$, $CE : EA = \triangle BOC : \triangle BOA$, and $AF : FB = \triangle COA : \triangle COB$. Compounding the three ratios cancels the three areas identically, yielding $+1$. Conversely, if $AD$ and $BE$ meet at $O$ and $CO$ meets $AB$ in $F'$, then $AF' : F'B = AF : FB$, so $F'$ coincides with $F$. Q.E.D.
Book VI, Section I, Proposition 4 (:person:`Menelaus’s <menelaus>` Theorem). If a transversal straight line cut the three sides $BC, CA, AB$ of a triangle (produced as necessary) in $D, E, F$, then with directed segments
$$ \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -1, $$
and conversely, if three points $D, E, F$ on the sides satisfy this signed product, they are collinear.
Book VI, Section II — Centres & Axes of Similitude¶
Definition (:term:`Centres of Similitude <centre-of-similitude>`). For any two circles $(O_1, r_1)$ and $(O_2, r_2)$, the points $S_e$ and $S_i$ dividing the line of centres $O_1 O_2$ externally and internally in the ratio $r_1 : r_2$ are called the External (Direct) and Internal (Transverse) Centres of Similitude. Every pair of parallel radii joins through $S_e$ (same sense) or $S_i$ (opposite sense), and all common tangents pass through $S_e$ or $S_i$.
Proposition 4 (Circle of Similitude). The locus of a point at which two given unequal circles subtend equal angles is the circle having the segment $S_e S_i$ joining their external and internal centres of similitude as diameter—coaxal with the two given circles!
Book VI, Section II, Proposition 6 (:person:`Monge’s <monge>` Theorem). Given three coplanar circles, their six centres of similitude lie three by three upon four straight lines—called the four Axes of Similitude:
all three external centres of similitude are collinear on the External Axis of Similitude;
any two internal centres of similitude are collinear with the third external centre of similitude.
Demonstration. Apply Menelaus's Theorem to the triangle of centres $\triangle O_1 O_2 O_3$ with radii $r_1, r_2, r_3$: the three external centres divide the sides externally in ratios $r_1/r_2, r_2/r_3, r_3/r_1$ whose product is $+1$ in magnitude and $(-1)^3 = -1$ in directed sign; replacing any two external divisions by internal divisions preserves the sign $(-1)(+1)(+1) = -1$. Q.E.D.
Book VI, Section III — Harmonic Section & the Complete Quadrilateral¶
Definition (:term:`Harmonic Section <harmonic-section>`). A segment $AB$ is divided harmonically at $C$ and $D$ when it is divided internally at $C$ and externally at $D$ in the same ratio ($AC : CB = AD : BD$), so that $AB$ is the harmonic mean of $AC$ and $AD$:
$$ \frac{2}{AB} = \frac{1}{AC} + \frac{1}{AD}, \qquad \text{or, with midpoint } O \text{ of } AB, \qquad OC \cdot OD = OA^2. $$
(Note the bridge to Unit 2: two points divide a diameter harmonically if and only if they are inverse points with respect to the circle on that diameter!)
Book VI, Section III, Proposition 7 (Harmonic Property of the :term:`Complete Quadrilateral <complete-quadrilateral>`). In any complete quadrilateral formed by four straight lines [AB], [BC], [CD], [DA] whose opposite sides meet at E and F, each of the three diagonals [AC], [BD], [EF] is divided harmonically by the other two:
$$ (A, C; P, Q_a) = -1, \qquad (B, D; P, Q_b) = -1, \qquad (E, F; Q_a, Q_b) = -1. $$
Demonstration. In $\triangle FAB$, the points D on $FA$ and C on $FB$ determine two cevians [AC] and [BD] concurring at P, while the line [CD] is a transversal meeting $FB$ at $C$, $FA$ at $D$, and $AB$ at E. Let the third cevian $FP$ meet $AB$ at $K$. Applying Ceva’s Theorem to $\triangle FAB$ with point $P$ gives $\frac{AK}{KB}\cdot\frac{BC}{CF}\cdot\frac{FD}{DA} = +1$, whereas Menelaus’s Theorem applied to $\triangle FAB$ with transversal $CDE$ gives $\frac{AE}{EB}\cdot\frac{BC}{CF}\cdot\frac{FD}{DA} = -1$. Dividing the two equations yields $(A, B; K, E) = -1$, so the pencil $F(A, B; P, E)$ is harmonic. Cutting this harmonic pencil by transversal [AC] gives $(A, C; P, Q_a) = -1$, cutting by [BD] gives $(B, D; P, Q_b) = -1$, and projecting from $A$ onto [EF] gives $(E, F; Q_a, Q_b) = -1$. Q.E.D.
Book VI, Section VI — Theory of Anharmonic Section, Desargues & Pascal¶
Definition (:term:`Anharmonic Ratio <anharmonic-ratio>`, :person:`Chasles <chasles>`). For four collinear points $A, B, C, D$, the anharmonic (cross) ratio $(ABCD) = \frac{AC}{CB} : \frac{AD}{DB} = \frac{AC \cdot BD}{AD \cdot BC}$ is invariant under central projection from any vertex $O$:
$$ (ABCD) = \frac{\sin \angle AOC \cdot \sin \angle BOD}{\sin \angle AOD \cdot \sin \angle BOC} = O(ABCD). $$
Proposition 5 (Anharmonic Property of Four Concyclic Points). If $A, B, C, D$ be four fixed points on a circle and $P$ any variable point on the circumference, the pencil $P(ABCD)$ has a constant anharmonic ratio independent of the position of $P$ (since equal arcs subtend equal inscribed angles by Euclid III.21).
Book VI, Section VI, Propositions 9–11 (:person:`Desargues’s <desargues>` Homology Theorem & :person:`Pascal’s <pascal>` Hexagrammum Mysticum).
Desargues’s Theorem (Coperspective $\iff$ Coaxial Triangles): If two triangles $\triangle ABC$ and $\triangle A'B'C'$ be in perspective so that $AA', BB', CC'$ concur at a Centre of Homology $O$, then the three intersections of corresponding sides $L = BC \cap B'C'$, $M = CA \cap C'A'$, $N = AB \cap A'B'$ lie on a single straight line (the Axis of Homology).
Pascal’s Mystic Hexagram (1640): If any hexagon $ABCDEF$ be inscribed in a circle, the three intersections of its pairs of opposite sides $L = AB \cap DE$, $M = BC \cap EF$, $N = CD \cap FA$ are collinear on the Pascal Line.
Demonstration of Pascal’s Theorem by Anharmonic Pencils. Project the four concyclic vertices $A, C, D, E$ from the remaining two vertices $B$ and $F$ on the circle; by Proposition 5, the two pencils have equal anharmonic ratios: $B(ACDE) = F(ACDE)$. Cut the pencil $B(ACDE)$ by transversal $DE$ to obtain the range $(L, K, D, E)$ where $L = BA \cap DE$ and $K = BC \cap DE$; cut the pencil $F(ACDE)$ by transversal $CD$ to obtain the range $(N, C, D, J)$ where $N = FA \cap CD$ and $J = FE \cap CD$. Because $(L, K, D, E) = (N, C, D, J)$ and the intersection $D = DE \cap CD$ corresponds to itself in both ranges, by Chasles’s perspective lemma (Prop. 4 Cor.) the joins of the remaining corresponding pairs—$LN$, $KC$ (line $BC$), and $EJ$ (line $EF$)—concur at a single point, namely $M = BC \cap EF$. Therefore $L, M, N$ are collinear. Q.E.D.