Let ABC be an isosceles triangle and D a point on its circumcircle (c), on the arc opposite to A. Draw the tangent tA to A and lines DB, DC intersecting that tangent respectively at points F and G. Then: [1] The circle (c') through points {D,F,G} is tangent to (c) at D. [2] Extend line DA to intersect (c') at E and draw the tangent there. Take its intersection points {L,E,N,M,K} with lines {DC,DA,DB,tD, JD}, where tD the (common to c, c') tangent at D and J the symmetric of A on its center. Then (L,N,E,K) = -1 is a harmonic division. Point M is the middle of EK. [3] The intersection points {Q,P,R} of pairs of lines {AC,BD}, {AB,CD}, {tA,tD} respectively, are collinear.
[1] Show that EFG is homothetic to ABC with respect to D. [2] DE/DK is interior/exterior bisector at D of triangle DBC, hence ( Apollonian_Circles.html ) (L,N,E,K) = -1. [3] Is a consequence of [2] (ibid). [4] Is Pascal's theorem applied to the cyclic quadrangle ABDC (see PascalOnQuadrangles.html ).
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