= = the sides EF, FD, DE of DEF. the perpendiculars AX, BY, CZ. = the segments AH, BH, CH of the perpendiculars. = the segments HX, HY, HZ of the perpendiculars. = = the perpendiculars OA', OB', OC' from the circumcentre. the internal angular bisectors of A, B, C. = the external angular bisectors of A, B, C. = radius of the incircle. radii of the 1st, 2nd, 3rd excircles. R R1, R2, R3 R1, R2', R2' ρ P1 P2 P3 S S1, S2, S3 X, Y, Z = radius of the incircle of XYZ, = radii of the 1st, 2nd, 3rd excircles of XYZ. Δ, Δε Δε Δε = AREAS. ABC, HCB, CHA, BAH. = III, III, I, II, I, II. § 1. CENTROID. The medians of a triangle are concurrent.* FIGURE 1. Let the medians BB', CC' cut each other at G; join AG, and let it cut BC at A'. Produce AA' to L, making GL equal to GA, and join BL, CL. Because DEF. The point G is called sometimes the centre of gravity † of the triangle ABC; sometimes the centre of mean distances of the points A, B, C; and more frequently now the centroid § of the triangle ABC. The simplest construction for obtaining G by means of the ruler and the compasses is the following || : With B as centre and AC as radius describe a circle; with C as centre and AB as radius describe a second circle cutting the former below the base at D. Join DC and produce it to meet the second circle at E. AD and BE intersect at the centroid G. (1) A'G={AG=}AA'. Hence the centroid of a triangle may be found by drawing any median and trisecting it; and if two (or a series of) triangles have the same vertex and the same median drawn from that vertex, they have the same centroid. * Archimedes, De planorum æquilibriis, I. 13, 14. + Archimedes. ‡ Carnot, Géométrie de Position, p. 315 (1803), and Lhuilier, Élémens d'Analyse, p. X. (1809). § This expression was suggested by T. S. Davies in 1843 in the Mathematician I. 58. It had been used by Dr Hey in 1814 to designate another point. || Mr E. Lemoine in the Report (second part) of the 21st session of the Asso ciation Française pour l'avancement des sciences, p. 77 (1892). (2) Triangle GBC=GCA=GAB=}ABC. (3) The sides of triangle A'B'C' are respectively parallel to those of ABC; hence these triangles are directly similar. Also, since the lines AA', BB', CC' joining corresponding vertices are concurrent at G, triangles ABC, A'B'C' are homothetic, and G is their homothetic centre. DEF.-Triangles such as the fundamental triangle ABC, and that formed by joining the feet of its medians have in recent years received the following names : A'B'C' is the complementary triangle of ABC. A'B'C'. These names are applied also to corresponding points* in such triangles. Thus if P be any point in or outside of triangle ABC, and P' be the corresponding point in or outside of triangle A'B'C', P' is the complementary point of P, Р ,, anticomplementary,, P'. (4) If А ̧3 ̧Ñ1 be the triangle formed by drawing through A, B, C parallels to the opposite sides of triangle ABC, ABC is the complementary triangle of A,B,C1, FIGURE 2. ABC. (5) The fundamental triangle ABC is directly similar to the triangles cut off from it by the sides of its complementary triangle, AC'B', C'BA', B'A'C. (6) The centroid of the fundamental triangle is the centroid of the complementary triangle; the centroid of the complementary triangle is the centroid of its complementary triangle; and so on. (7) All straight lines parallel to the base of a triangle and terminated by the other sides are bisected by the median to the base. * See Mr Emile Vigarié's articles Sur les Points Complémentaires in Mathesis, VII., 5-12, 57-62, 84-9, 105-110 (1887). |