[alogo] Three collinear points

Consider a triangle t = ABC inscribed in a conic (c), and a line (e) intersecting this conic at points D, E. Let A*, B*, C* be the intersection points of the sides of the triangle BC, CA, AB with (e). Let further H be the intersection point of BB*, AA*, F be the intersection point of CC* and (c) and I the harmonic conjugate of C* w.r. to D,E. Then F, I, H are on a line. (This property generalizes the one discussed in ThreeCollinearPts.html )

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

1) Consider the quadrilateral AG''BG'. The intersection point I of the diagonals G'G'', FH is harmonic conjugate to C''', with respect to F,H, C''' being the intersection point of AB, FH. But since (C*A, C*B*, C*C, C*H) is a harmonic bundle, the same is true for the intersection point of FH with CC*. Hence FH, G'I and CC* intersect at a point I.
2) It follows that (AC*, AI, AC', AH) is a harmonic bundle, hence (C*IA*C')=-1. This implies that (GA*, GC', GI, GC*) is a harmonic bundle of lines hence line GI is the polar of C*, proving the collinearity.

Note that in the case D, E coincide i.e. (e) is tangent to (c), then I is the contact point of (c) and (e) and this is collinear with F and H. This property is used in the construction of the family of conics through four points and tangent to a given line, discussed in the file AllConicsCircumscribed.html .

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