[alogo] Polar of a point with respect to a circle

Consider a circle (c) a point A and two lines (a) , (b) through A, intersecting (c) at B, D and C, E respectively. Show that Line d = [FG], joining the intersection points of CB, DE and CD, BE respectively, is the [polar] of point A with respect to point C. This line is characterized by the fact that it contains all points I on secants (b) through A, such that A, I are harmonic conjugate to the intersection points C, E of circle (c) with the line.

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

The result follows immediately from the discussion in Polar.html . Indeed, there we examined the special case where line (a) coincides with the diameter line [AKL] through A and proved that line MH, H being the projection of M on AL, has the characteristic property of the polar.

F, G are on this line MH. Indeed, per definition, the polar defines on a, b points I, J which are respectively harmonic conjugate to A, with respect to C, E and BD respectively. But the same does the line FG on lines a and b (see Harmonic.html ). Thus, I, J being uniquely determined on b, a respectively, FG coincides with MH.

Various interesting facts are related to this figure.
1) The intersection point G of the diagonals CD, BE of the cyclic quadrilateral is the pole of line AF. In fact, by the previous arguments G is on the polar of A and analogously it is on the polar of F. But the pole-polar relation has a duality behaviour. If point A is on the polar of point B then the polar of A contains B. This identifies AF with the polar of G. Thus the orthogonal to AF through G passes through the center O of circle c.
2) The inversion with respect to the circle f(A, |AS|), AS being a tangent of (c) from A preserves the lines through A and the circle (c), c being orthogonal to f. By this inversion circle g, with diameter AF is mapped to line h=[GO]. Hence P, a intersection point of h and g remains fixed by the inversion and lies on the inversion circle f.
3) Circles {PCE} and {PBD} are tangent at P, their common tangent being line AP. In fact, C, E are inverse points with respect to the inversion on f. Hence circle {PCE} remains invariant by this inversion, which leaves P fixed and is orthogonal to f, having tangent there the line PA, its center being on PF. The same arguments apply to {PBD}. These remarks are used in the discussion of orthodiagonal quadrilaterals contained in the file OrthodiagonalFromCyclic.html .

Analogously one shows that circles {PBC} and {PED} are tangent at p, their common tangent being line PF. Look at the file FourTangentCircles.html for the continuation of the subject.

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