R being each 0° and Q=180°. The species of the limiting triangle is determined by the ratios QR: RP: PQ, or their equivalents a. AO:b. BO: c. CO. (Art. 23, Ex. 8.) Hence if a transversal is drawn to a triangle such that the ratios of its segments made by the sides is constant; the ratios AO: BO: CO are known and with them the point O. As in the general case, the triangles QOR, ROP, POQ are constant in species. It follows then that if P, Q, R be the feet of the perpendiculars from O on the sides of ABC, and the lines OP, OQ, OR rotated through any angle in the same direction, P, Q, R will always remain collinear and the ratios PQ: QR: RP are constant.* COR. Ptolemy's Theorem.-Since QR: RP: PQ=a. AO: b. BO: c. CO, and PQ+QR=PR; therefore a. AO+c. CO=b. BO. EXAMPLES. 1. Place a given line PQ divided in any point R such that the points P, Q, R may lie in an assigned order on the sides of a given triangle. * Chasles' Géométrie supérieure, p. 281. 2. Draw a line across a quadrilateral, meeting the sides in PQRS such that the ratios PQ: QR: RS may be given. 3. The line joining 0 to the orthocentre of ABC is bisected by the Simson line PQR, and intersects it on the nine points circle. 4. The angle subtended by any two points O, and 0, on the circle is equal to the angle between their Simson lines. 5. The Simson lines of two points diametrically opposite intersect at right angles on the nine points circle. (By Ex. 4.) 25. Theorem. For three positions, PQR, PQR1, PQR2 of the triangle of given species inscribed in a given one ABC; to prove that PP1: PP2=QQ1: QQ2 = RR1: RR2. 1 Since the triangles OPQ, OP,Q1, OP2Q2 are similar, we have OP: OP=OQ: 0Q1, also POP1 =QOQ, since POQ=POQ; therefore the triangles POP, and QOQ, are similar. Hence Similarly therefore Similarly therefore, etc. PP1:QQ1=OP: 0Q. PP1:QQ1: RR1 = OP: OQ: OR. Now if PQR and P2Q2R2 denote two fixed positions of the variable inscribed triangle PQR of constant species, and PQR any arbitrary position, it follows that a variable line PQ, dividing similarly two linear segments P1Q1 and PQ subtends a constant angle POQ at a fixed point O. The point O is determined by the intersection of the loci of the vertices of the triangles PQ10 and P20, whose bases PQ1 and PQ, are given and ratio of sides (=P1P2: QQ1), or the intersection of the circles CP1Q1 and CP2Q2 1 2 Since PP and QQ, form similar triangles with O, this point is termed the Centre of Similitude of the segments. Thus the centre of similitude of two segments AB and CD is the intersection of the circles passing through the two pairs of non-corresponding extremities and the intersection O of the given lines. Or it may be regarded as the common vertex of two similar triangles described on the sides. If the points B and D coincide, O coincides with them, and the circle ADO meeting CD in coincident points D and O therefore touches CD. In the same case the circle BCO touches AB. COR. The centres of similitude of the sides of a triangle taken in pairs are therefore found by describing circles on BC and AC touching the sides AC and BC respectively. The second point of intersection of these circles is a centre of similitude of AC and BC; similarly for each of the remaining pairs of sides. EXAMPLES. 1. Draw a line L dividing three linear segments A142, B1B2 and C1C2 in the same ratio. (Dublin Univ. Exam. Papers.) [Let the required line intersect the segments in P, Q and R respectively, O, and O2 the centres of similitude of the pairs of lines A14 BB, and BB, CC. Then in the triangle 0,202 we know the base 002 and vertical angle, since it is equal to 180-0 QP-02QR; therefore, etc.] 2. The centres of similitude of the sides of a triangle taken in pairs are the middle points of the symmedian chords of the circumcircle. [Let X, Y, Z denote the middle points of the sides of the triangle ABC; CD and CE the median and symmedian chords of the circle respectively; M the middle point of CE. Join ZE, AM and BM. Then since LACD=BCE and LCAZ CEB, the triangles ACZ and ECB are similar, and Y and M being the middle points of a pair of corresponding sides, CYZ and CMB are therefore similar. Hence LCBM CZY=BCZ-ACM. Similarly LCAM-BCM; therefore the triangles BCM and CAM are similar.] 3. Prove the following results from Ex. 2: 1°. CEZ-difference of base angles (B-A). 2°. The triangles ADZ and BEZ equal in every respect. 4°. CM=aba2+b2+2ab cos C. 5°. BMC CMA=π-C. = 6°. The circum-circle of ABM passes through the centre of the circle ABC. 4. Having given the base (c) bisector of base (CZ) and difference of base angles (B-A); construct the triangle. [The triangle CEZ is readily constructed; therefore, etc.] 5. Having given the bisector of base (CZ) rectangle under sides (ab) and difference of base angles (B-A); construct the triangle. [As in Ex. 4.] 6. Having given the base, median, and symmedian of a triangle; construct it. SECTION IV. MISCELLANEOUS PROPOSITIONS. 26. Prop. I.-Through a point P to draw a line across an angle such that the intercepted segment MN may subtend at a fixed point Q a triangle of maximum area. |