= where 28 CC'. It evidently passes through the intersections of the given circles.] 12. Show that CM. C'N=const. [Draw CX at right angles to C'N. Join OX. Since OC'X is an isosceles triangle and N a point in the base produced, CM. C'N C'N. NX=ON2 - OX2=ON2-OC'2 where ✪ is the angle between the given circles; therefore, etc.] 13. Any circle described around the polar centre of a triangle ABC meets the corresponding sides of the median triangle in A', B', C' such that AA'=BB' = CC'. 14. A tangent is drawn from the polar centre to the circumcircle, and from the point of contact a tangent is drawn to the polar circle, show that the angle between these lines is 45°. 15. Draw through P a line cutting each of two given circles in conjugate points with respect to the other. [By Exs. 10 and 11.] 16. Draw a line cutting each of two circles X and Y in conjugate points with respect to a third (Z). [Let the required line meet Z in the points A and B. The middle point M of AB is the intersection of two known circles passing through the intersections of Z and X and Z and Y (Ex. 11), and is thus determined; therefore, etc.] SECTION II. 79. Salmon's Theorem.-The distances of any two points A and B from the centre O of a circle are proportional to the distances AM and BL of each from the polar of the other. Draw AB' and BA' perpendiculars to OB and OA respectively. Then OA. OL-OB. OM=2, and since AA'BB' is a cyclic quadrilateral, OA. OA'=OB. OB'; therefore OB' OM OM-OB' B'M AM OA = OL therefore, etc. By alternation OA/AM=OB/BL. COR. 1. If M is a fixed line and OA/AM a constant ratio, B is a fixed point and the envelope of L is a circle; or, the pole of a variable tangent to a circle with respect to another given circle is such that its distance from the centre of the latter bears a fixed ratio to the distance from a fixed line. COR. 2. If A and B are both on the circle (0, r); OA = OB, and therefore AM= BL; or, the points of contact of tangents to a circle are equidistant from the tangents as is otherwise evident (Euc. I. 26). COR. 3. Let B and its polar M vary and the different positions be denoted by B1, B2, B3, then since OA OB OA OB1 OA = = = M1, M2, M3, OB2 etc.; 3. ... by multiplication of ratios, we have OAn = OB. OB1. OB2...; 2 AM. AM ̧ . AM1⁄2 ..... ̄ ... or, the product of the distances of a point (A) from any number of lines (M) is to the product of the distances of their poles (B) from the polar (L) of the point as the nth power of the distance of the point from the centre is to the product of the distances of the poles from the centre. 2 COR. 4. If M, M1, M2 in Cor. 3 form an inscribed polygon, B, B1, B2, ... are the vertices of the corresponding escribed one; hence the product of the distances of any point from the sides of an in-polygon is to the product of the distances of the vertices of the corresponding ex-polygon from the polar of the point as the nth power of the distance of the latter from the centre is to the product of the distances of the vertices of the expolygon from the centre. COR. 5. The rectangle under the distances of the extremities of any chord from a tangent is equal to the square of the distance of its point of contact from the chord. EXAMPLES. 1. The opposite vertices of an escribed quadrilateral are AA', BB', CC'; to prove that OA. OA': OB. OB' : OC. OC'=AX. A'X : BX. B'X: CX. C'X, where X is a tangent to the circle at any point P. [Let the corresponding pairs of sides of the in-quadrilateral be L, L'; M, M'; N, N'; then since OA. OA' OP2 = multiplying these equations, AX.AX PL.PL' but PL. PL'=PM. PM'=PN. PN'; therefore, etc.] 2. If a, ẞ, y denote the perpendiculars from any point on the circum-circle on the sides of an in-triangle, or By sin A+ya sin B+aß sin C=0 α b с 3. If λ, μ, v be the perpendiculars from the vertices of any triangle upon a variable tangent to the in-circle, [Let A', B', C', P be the points of contact with the sides and any where a' is the perpendicular from P on B'C'. but OA. B'C' =2r2cot 4; substituting, we have A particular case of this has been noticed in Art. 55, Ex. 8.] 4. If the perpendiculars from the vertices on any tangent to the circum-circle of a triangle be λ, μ, v; to prove that a√λ+b√μ+c√/v=0. [If P be the point of contact of the tangent to the circle, by Ptolemy's Theorem, a. AP+b. BP+c.CP=0, but AP2=2rλ, etc., hence Zaλ=0.] 5. For any point Pon the in-circle whose distances from the sides are a, ẞ, y; to prove that cos 4a+cos BJB+cos C√y=0. [Let A', u', v' be the distances of the points of contact A', B', C of the sides of ABC from the tangent at P; a', B', y' the distances of P from the sides of A'B'C'. *The angles of A'B'C' are respectively 90-A, 90–†B, 90–1C; therefore a': b': c' costA: cos B: cos C. hence, on substituting, since a'=2r cos A, Za=0=cos A/a, therefore, etc.] NOTE.-The equations in Exs. 2 and 5 are known in Analytical Geometry to be those of the circum- and in-circles respectively, the given triangle ABC being taken as the triangle of reference. The expressions in Exs. 3 and 4 are the Tangential Equations of the Inand Circum-Circles. 6. If two triangles ABC, A'B'C' are reciprocal polars, they are in perspective. [Let the perpendiculars from A'B'C' on the sides of ABC be P1, P2, P3; 91, 92, 93; 71, 72, 73 respectively; then, by Salmon's Theorem, OC" 71 OA' Pa OB'_93 ; ос T2 ΟΑ' P3 OB' 91 multiplying these equations we have p2. P. 3. 11; therefore, etc. (Art. 65.)] 7. A triangle inscribed in a circle is in perspective with the corresponding escribed one. [By Ex. 6.] 8. Any two triangles may be so placed that the vertices of either are the poles of the sides of the other with respect to a circle. [At the centre O of the required circle the sides of each triangle subtend angles similar to those of the other triangle. Find points |