Unit 5: Supplementary Chapter, Secs. I & III (Part 1) — Symmedians, Lemoine’s Point & Lemoine Circles¶
John Casey — Supplementary Chapter on the Recent Elementary Geometry of the Triangle (5th ed., 1888 / 6th ed., 1892)
Section I — Isogonal & Isotomic Conjugates and Antiparallels¶
Definition (:term:`Isogonal Conjugates <isogonal-conjugates>`). Two straight lines drawn from a vertex of a triangle making equal angles with the internal bisector of that angle are called isogonal conjugates.
Supplementary Chapter, Section I, Proposition 1 (Isogonal Concurrency & Pedal Circle). If three lines drawn from the vertices $A, B, C$ of a triangle concur at a point $P$, their three isogonal conjugates also concur at a single point $P'$ (the Isogonal Conjugate of $P$*); and the six feet of perpendiculars let fall from* $P$ and $P'$ upon the three sides of $\triangle ABC$ lie upon a single circle whose centre is the midpoint of $PP'$.
Classical Specialization: The circumcentre O and orthocentre $H$ are isogonal conjugates, and their shared six-point pedal circle (the three side midpoints and three altitude feet) is the Nine-Points Circle centered at $N = \operatorname{midpoint}(O, H)$!
Definition (:term:`Isotomic Conjugates <isotomic-conjugates>`). Two points $D, D'$ on side $BC$ equidistant from its midpoint $M_a$ ($BD = D'C$) are called isotomic; by Ceva’s Theorem, if $AD, BE, CF$ concur at $P$, their isotomic cevians $AD', BE', CF'$ concur at the Isotomic Conjugate of $P$ (for instance, the Gergonne and Nagel points).
Definition (:term:`Antiparallels <antiparallel>`). A transversal cutting $AB$ and $AC$ at $D$ and $E$ so that $\angle ADE = \angle ACB$ and $\angle AED = \angle ABC$ is called antiparallel to $BC$ with respect to $\angle A$. Every antiparallel to $BC$ makes a cyclic quadrilateral $BCED$ with $BC$, and is parallel to the tangent to the circumcircle (O A) at A (by Euclid III.32).
Theory of Symmedians & Lemoine’s Point $K = X(6)$¶
Definition (:term:`Symmedians <symmedian>`). The isogonal conjugate of a median of a triangle is called a Symmedian.
Section I, Propositions 5–7 (:person:`Lemoine’s <lemoine>` Symmedian Point & Minimum-Square Property).
Concurrency at $K$: The three symmedians [A K], [B K], [C K] of △ABC concur at a single point K (the Symmedian Point or Lemoine’s Point, $X(6)$, isogonal conjugate of the Centroid $G$).
Antiparallel Bisection & Tangency Construction: Each symmedian [A K] bisects every antiparallel to $BC$, passes through the pole of $BC$ with respect to the circumcircle (O A) (the intersection of the circumcircle tangents at $B$ and $C$*), and divides the opposite side* $BC$ in the ratio of the squares of the adjacent sides ($c^2 : b^2$).
Proportional Distances & Lagrange’s Minimum: The perpendicular distances $(x, y, z)$ from K upon the sides $(a, b, c)$ are directly proportional to the lengths of those sides:
$$ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \frac{2\Delta}{a^2 + b^2 + c^2}, $$
and among all points in the plane, K strictly minimizes the sum of squares of perpendicular distances to the sides, $x^2 + y^2 + z^2 \ge \frac{4\Delta^2}{a^2 + b^2 + c^2}$.
Demonstration of Minimum Property. By Cauchy’s / Lagrange’s identity,
$$ (a^2 + b^2 + c^2)(x^2 + y^2 + z^2) - (ax + by + cz)^2 = (bx - ay)^2 + (cy - bz)^2 + (az - cx)^2 \ge 0. $$
Since $ax + by + cz = 2\Delta$ is constant for every interior point, $x^2 + y^2 + z^2$ attains its unique minimum $\frac{4\Delta^2}{a^2 + b^2 + c^2}$ if and only if $x : y : z = a : b : c$, which uniquely characterizes K. Q.E.D.
Section III (Part 1) — Lemoine’s First Circle & the Cosine Circle¶
Section III, Proposition 1 (:term:`Lemoine’s First Circle / Triplicate-Ratio Circle <lemoine-circle>`). If through the Symmedian Point K three straight lines [parBC], [parCA], [parAB] be drawn parallel respectively to the sides [BC], [CA], [AB], meeting the sides in six points D1, D2 on $BC$, E1, E2 on $CA$, and F1, F2 on $AB$:
All six points lie upon a single circle—Lemoine’s First Circle (Ml D1);
Its centre Ml ($X(182)$) is the midpoint of the segment joining the Circumcentre O and Symmedian Point K;
The three chords $|D_1 D_2|, |E_1 E_2|, |F_1 F_2|$ cut off upon the sides $BC, CA, AB$ are directly proportional to the cubes of those sides ($a^3 : b^3 : c^3$, hence :person:`Tucker’s <tucker>` name “Triplicate-Ratio Circle”).
Demonstration. In parallelogram $A F_2 K E_1$ ($K E_1 \parallel AB$ and $K F_2 \parallel AC$), the diagonal $AK$ lies along the symmedian [A K]. Because the adjacent sides of $A F_2 K E_1$ are proportional to the distances from $K$ to $AC, AB$ divided by $\sin A$—hence proportional to $b : c$—triangle $\triangle A F_2 E_1$ is inversely similar to $\triangle ABC$, so the chord $E_1 F_2$ is antiparallel to $BC$! Meanwhile $E_2 F_1$ passes through $K$ parallel to $BC$; hence $F_2 F_1 E_2 E_1$ is an isosceles-angle cyclic trapezium, making $E_1, E_2, F_1, F_2$ concyclic. By cyclic symmetry, $F_1, F_2, D_1, D_2$ and $D_1, D_2, E_1, E_2$ are each concyclic; since any two of these three circles share two points on one side and would otherwise have three non-concurrent radical chords $BC, CA, AB$, all six points $D_1, D_2, E_1, E_2, F_1, F_2$ lie on a single circle (Ml D1). Moreover, the perpendicular bisectors of the three antiparallel chords $D_2 E_1, E_2 F_1, F_2 D_1$ bisect segment $OK$, fixing the centre at Ml $= \operatorname{midpoint}(O, K)$. Finally, since $K = (a^2 : b^2 : c^2)$, the parallel intercepts cut off middle segment $|D_1 D_2| = \frac{a^3}{a^2 + b^2 + c^2}$, proportional to $a^3 : b^3 : c^3$. Q.E.D.
Proposition 1, Corollary 4 (:term:`The Cosine Circle / Second Lemoine Circle <cosine-circle>`). If through K three straight lines be drawn antiparallel respectively to $BC, CA, AB$, each is bisected at K and all three antiparallel segments have equal length! Consequently their six extremities on the sides of $\triangle ABC$ lie on a single circle (K Cb2) centered at K itself, cutting off chords on $BC, CA, AB$ proportional to $\cos A : \cos B : \cos C$ (hence called the Cosine Circle).