92 ; sin BA'X' 93 sin CA'X sin CAX=22, etc.; hence the above expression is equivalent to 9272 73P3 P191-91. 12. P3-1.] = = COR. 4. Brianchon's Theorem.-Let AC'BA'CB' be an escribed hexagon and x, y, z the intercepts made by the circle on the sides of the triangle A'B'C'; since sin BA'X sin CA'X_y2. sin BA'X'sin CA'X' = with two other similar equations, Cor. 3 in this particular *The property on which this depends is as follows:-If from the point of intersection C of two tangents CA, CB to a circle a secant of length x is drawn dividing the angle ACB into segments a and B; then sin a sin β α χ. For if O be the centre of the circle and OX a perpendicular to the secant, we have sin a sin ẞ=sin(a +ẞ) - sin2(a - ẞ) = r2/OC2 - OX2/002 = x2/4002; therefore, etc. case reduces to:-The lines connecting the opposite vertices of an escribed hexagon are concurrent; or, the two triangles formed by joining the alternate vertices of an escribed hexagon are in perspective. The centre and axis of perspective of the triangles are termed the Brianchon* Point and Line of the hexagon AC'BA'CB', which for the same reason is called a Brianchon Hexagon. COR. 5. If two of the sides AF and EF of an escribed hexagon coincide, the vertex F is the point of contact of the tangent AE (Art. 6); hence for an escribed pentagon ABCDE, if the lines AD and BE meet in O, the points C, O, F are collinear (cf. Art. 63, foot-note). COR. 6. If two pairs of sides BC, CD and AF, EF coincide, the hexagon reduces to a quadrilateral ABDE; hence the diagonals AD and BE meet on CF; similarly they meet on C'F; therefore the internal diagonals of an escribed quadrilateral. and of the corresponding inscribed meet in a point. * Published by Brianchon in the year 1806, and derived by him from Pascal's Theorem by the process of reciprocation with respect to the circle. (See Art. 80, 2°.) COR. 7. Consider the cyclic hexagon FFC'CCF. Its Pascal line is the line of collinearity of the three points (1) FF, CC; (2) FC, CF'; (3) FF, CC: but the line joining (2) and (3) is the third diagonal of the inscribed quadrilateral CFCF" and (1) is the intersection of the tangents at C and F, and therefore one extremity of the third diagonal of the escribed quadrilateral; hence :-the third diagonals of any inscribed and corresponding escribed quadrilaterals coincide. COR. 8. Let PQRS be any cyclic quadrilateral; and let XX'YY'ZZ', the corresponding escribed quadrilateral, be regarded as a Brianchon hexagon ZPX'Z'RX whose two pairs of coincident sides are the tangents from Y. Then the lines ZZ', PR, XX' are concurrent at the Brianchon point B; similarly, if the pairs of coincident sides are the tangents from Y', we have ZZ', QS, XX concurrent, i.e. the pairs of opposite connectors PR and QS of the inscribed quadrilateral and ZZ' and XX' of the corresponding escribed cointersect. We see therefore from Cors. 7 and 8 that any pair of opposite connectors of an inscribed quadrilateral and the corresponding pair for the quadrilateral escribed at its vertices are concurrent. The three points of concurrence on the figure are A, B, C. The points U, V, W, U', V', W' lie in triads on four lines. EXAMPLES. 1. Three pairs of tangents are drawn from the vertices of a triangle to any circle to meet the opposite sides in points XX', YY', ZZ'; show that if X, Y, Z are collinear, X, Y, Z are also collinear. [Apply Cor. 4.] 2. ABC is a triangle inscribed in and in perspective with A'B'C'; the tangents from ABC to the in-circle of A'B'C' meet the opposite sides in three collinear points X, Y, Z (BC in X, etc.). [Let the axis of perspective of the two triangles be X'Y'Z', therefore by Cor. 4 we have (B etc., by Ex. 1.] 3. If points XX', YY', ZZ' be taken on the sides of a triangle BX BX' CY CY' AZ AZ' such that CX' CX'' AY AY'' BZ' BZ' they are the vertices of a Pascal hexagon. = = 1, 4. The lines joining each pair of points to the opposite vertex (AX and AX', etc.) of the triangle determine a Brianchon hexagon. 5. (a) Any two transversals XYZ, X'Y'Z' determine on the sides the vertices of a Pascal hexagon. (B) Two triads of points on the sides which connect concurrently with the opposite vertices determine a Pascal hexagon. (y) A transversal XYZ and three points X', Y', Z' which connect concurrently with the opposite vertices determine a Brianchon hexagon. 6. A hexagon is inscribed in a circle; prove that the continued products of the perpendiculars from any point on the Pascal line on the alternate sides are equal (xyz=x'y'z). [Let AB'CA'BC' be the hexagon whose pairs of opposite sides BC', BC; CA', C'A; AB, A'B meet in points X, Y, Z respectively and the Pascal line L (XYZ) at angles a, a', B, B', y, y'; then Similarly, BL.CL. BX. C'X sin'a sin2a B'L. CL B'X. CX sin'a' = and sin2a" = C'L. AL_sin2B AL. BL siny Multiplying these equations and reducing, (Euc. III. 36) sin'a sin2ß sin3y=sin a'sin'B'sin'y'; therefore, etc.] 7. From the middle points L, M, N of the sides of a triangle tangents are drawn to the in-circle; show that these tangents form a triangle (A'B'C') in perspective with that (PQR) obtained by joining the points of contact of the in- or ex-circles with the sides, and the centre of perspective is the median point of ABC. [For since the sides of ABC with any two of the tangents form |